Mercurial > sdl-ios-xcode
annotate src/video/e_sqrt.h @ 4158:96ce26f24b01 SDL-1.2
Date: Sun, 7 Sep 2008 15:17:00 +0200
From: c2woody@gmx.net
Subject: [SDL] SDL 1.2 doube free/pointer zeroing missing
Hello,
this is about a crash/debug breakage for the current SDL 1.2
source tree (today's svn checkout, same problem in 1.2.13 and
before as far as relevant).
In some places memory is free()d but the associated pointer
is not zeroed, leading to for example double free()s.
For me this happened because SDL_StopEventThread() was executed
twice (during restart of the subsystems), once for the close
down in SDL_VideoQuit() and once at the startup, right at the
beginning of SDL_StartEventLoop(). Thus the code
SDL_DestroyMutex(SDL_EventQ.lock);
(see SDL_events.c) was called twice and executed the SDL_free(mutex);
twice as well, leading to a crash (msvc 64bit for which it was noticed).
I've tried to check all other occurrences of SDL_free and similar
code in msvc, see the attached patch (udiff against revision 4082).
Non-windows only codepaths have neither been checked nor touched.
Comments/ideas welcome.
Attached patch: NULLifies some pointers after they have been free()d.
author | Sam Lantinga <slouken@libsdl.org> |
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date | Wed, 12 Nov 2008 17:23:40 +0000 |
parents | 7a610f25c12f |
children | 782fd950bd46 c121d94672cb |
rev | line source |
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1 /* @(#)e_sqrt.c 5.1 93/09/24 */ |
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2 /* |
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3 * ==================================================== |
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4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. |
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5 * |
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6 * Developed at SunPro, a Sun Microsystems, Inc. business. |
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7 * Permission to use, copy, modify, and distribute this |
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8 * software is freely granted, provided that this notice |
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9 * is preserved. |
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10 * ==================================================== |
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11 */ |
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12 |
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13 #if defined(LIBM_SCCS) && !defined(lint) |
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14 static char rcsid[] = "$NetBSD: e_sqrt.c,v 1.8 1995/05/10 20:46:17 jtc Exp $"; |
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15 #endif |
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16 |
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17 /* __ieee754_sqrt(x) |
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18 * Return correctly rounded sqrt. |
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19 * ------------------------------------------ |
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20 * | Use the hardware sqrt if you have one | |
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21 * ------------------------------------------ |
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22 * Method: |
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23 * Bit by bit method using integer arithmetic. (Slow, but portable) |
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24 * 1. Normalization |
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25 * Scale x to y in [1,4) with even powers of 2: |
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26 * find an integer k such that 1 <= (y=x*2^(2k)) < 4, then |
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27 * sqrt(x) = 2^k * sqrt(y) |
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28 * 2. Bit by bit computation |
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29 * Let q = sqrt(y) truncated to i bit after binary point (q = 1), |
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30 * i 0 |
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31 * i+1 2 |
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32 * s = 2*q , and y = 2 * ( y - q ). (1) |
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33 * i i i i |
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34 * |
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35 * To compute q from q , one checks whether |
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36 * i+1 i |
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37 * |
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38 * -(i+1) 2 |
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39 * (q + 2 ) <= y. (2) |
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40 * i |
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41 * -(i+1) |
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42 * If (2) is false, then q = q ; otherwise q = q + 2 . |
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43 * i+1 i i+1 i |
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44 * |
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45 * With some algebric manipulation, it is not difficult to see |
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46 * that (2) is equivalent to |
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47 * -(i+1) |
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48 * s + 2 <= y (3) |
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49 * i i |
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50 * |
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51 * The advantage of (3) is that s and y can be computed by |
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52 * i i |
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53 * the following recurrence formula: |
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54 * if (3) is false |
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55 * |
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56 * s = s , y = y ; (4) |
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57 * i+1 i i+1 i |
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58 * |
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59 * otherwise, |
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60 * -i -(i+1) |
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61 * s = s + 2 , y = y - s - 2 (5) |
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62 * i+1 i i+1 i i |
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63 * |
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64 * One may easily use induction to prove (4) and (5). |
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65 * Note. Since the left hand side of (3) contain only i+2 bits, |
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66 * it does not necessary to do a full (53-bit) comparison |
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67 * in (3). |
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68 * 3. Final rounding |
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69 * After generating the 53 bits result, we compute one more bit. |
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70 * Together with the remainder, we can decide whether the |
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71 * result is exact, bigger than 1/2ulp, or less than 1/2ulp |
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72 * (it will never equal to 1/2ulp). |
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73 * The rounding mode can be detected by checking whether |
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74 * huge + tiny is equal to huge, and whether huge - tiny is |
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75 * equal to huge for some floating point number "huge" and "tiny". |
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76 * |
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77 * Special cases: |
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78 * sqrt(+-0) = +-0 ... exact |
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79 * sqrt(inf) = inf |
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80 * sqrt(-ve) = NaN ... with invalid signal |
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81 * sqrt(NaN) = NaN ... with invalid signal for signaling NaN |
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82 * |
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83 * Other methods : see the appended file at the end of the program below. |
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84 *--------------- |
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85 */ |
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86 |
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87 /*#include "math.h"*/ |
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88 #include "math_private.h" |
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89 |
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90 #ifdef __STDC__ |
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91 double SDL_NAME(copysign)(double x, double y) |
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92 #else |
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93 double SDL_NAME(copysign)(x,y) |
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94 double x,y; |
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95 #endif |
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96 { |
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97 u_int32_t hx,hy; |
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98 GET_HIGH_WORD(hx,x); |
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99 GET_HIGH_WORD(hy,y); |
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100 SET_HIGH_WORD(x,(hx&0x7fffffff)|(hy&0x80000000)); |
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101 return x; |
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102 } |
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103 |
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104 #ifdef __STDC__ |
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105 double SDL_NAME(scalbn) (double x, int n) |
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106 #else |
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107 double SDL_NAME(scalbn) (x,n) |
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108 double x; int n; |
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109 #endif |
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110 { |
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111 int32_t k,hx,lx; |
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112 EXTRACT_WORDS(hx,lx,x); |
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113 k = (hx&0x7ff00000)>>20; /* extract exponent */ |
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114 if (k==0) { /* 0 or subnormal x */ |
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115 if ((lx|(hx&0x7fffffff))==0) return x; /* +-0 */ |
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116 x *= two54; |
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117 GET_HIGH_WORD(hx,x); |
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118 k = ((hx&0x7ff00000)>>20) - 54; |
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119 if (n< -50000) return tiny*x; /*underflow*/ |
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120 } |
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121 if (k==0x7ff) return x+x; /* NaN or Inf */ |
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122 k = k+n; |
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123 if (k > 0x7fe) return huge*SDL_NAME(copysign)(huge,x); /* overflow */ |
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124 if (k > 0) /* normal result */ |
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125 {SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); return x;} |
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126 if (k <= -54) { |
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127 if (n > 50000) /* in case integer overflow in n+k */ |
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128 return huge*SDL_NAME(copysign)(huge,x); /*overflow*/ |
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129 else return tiny*SDL_NAME(copysign)(tiny,x); /*underflow*/ |
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130 } |
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131 k += 54; /* subnormal result */ |
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132 SET_HIGH_WORD(x,(hx&0x800fffff)|(k<<20)); |
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133 return x*twom54; |
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134 } |
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135 |
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136 #ifdef __STDC__ |
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137 double __ieee754_sqrt(double x) |
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138 #else |
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139 double __ieee754_sqrt(x) |
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140 double x; |
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141 #endif |
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142 { |
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143 double z; |
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144 int32_t sign = (int)0x80000000; |
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145 int32_t ix0,s0,q,m,t,i; |
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146 u_int32_t r,t1,s1,ix1,q1; |
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147 |
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148 EXTRACT_WORDS(ix0,ix1,x); |
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149 |
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150 /* take care of Inf and NaN */ |
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151 if((ix0&0x7ff00000)==0x7ff00000) { |
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152 return x*x+x; /* sqrt(NaN)=NaN, sqrt(+inf)=+inf |
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153 sqrt(-inf)=sNaN */ |
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154 } |
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155 /* take care of zero */ |
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156 if(ix0<=0) { |
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157 if(((ix0&(~sign))|ix1)==0) return x;/* sqrt(+-0) = +-0 */ |
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158 else if(ix0<0) |
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159 return (x-x)/(x-x); /* sqrt(-ve) = sNaN */ |
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160 } |
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161 /* normalize x */ |
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162 m = (ix0>>20); |
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163 if(m==0) { /* subnormal x */ |
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164 while(ix0==0) { |
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165 m -= 21; |
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166 ix0 |= (ix1>>11); ix1 <<= 21; |
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167 } |
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168 for(i=0;(ix0&0x00100000)==0;i++) ix0<<=1; |
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169 m -= i-1; |
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170 ix0 |= (ix1>>(32-i)); |
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171 ix1 <<= i; |
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172 } |
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173 m -= 1023; /* unbias exponent */ |
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174 ix0 = (ix0&0x000fffff)|0x00100000; |
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175 if(m&1){ /* odd m, double x to make it even */ |
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176 ix0 += ix0 + ((ix1&sign)>>31); |
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177 ix1 += ix1; |
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178 } |
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179 m >>= 1; /* m = [m/2] */ |
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180 |
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181 /* generate sqrt(x) bit by bit */ |
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182 ix0 += ix0 + ((ix1&sign)>>31); |
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183 ix1 += ix1; |
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184 q = q1 = s0 = s1 = 0; /* [q,q1] = sqrt(x) */ |
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185 r = 0x00200000; /* r = moving bit from right to left */ |
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186 |
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187 while(r!=0) { |
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188 t = s0+r; |
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189 if(t<=ix0) { |
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190 s0 = t+r; |
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191 ix0 -= t; |
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192 q += r; |
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193 } |
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194 ix0 += ix0 + ((ix1&sign)>>31); |
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195 ix1 += ix1; |
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196 r>>=1; |
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197 } |
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198 |
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199 r = sign; |
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200 while(r!=0) { |
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201 t1 = s1+r; |
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202 t = s0; |
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203 if((t<ix0)||((t==ix0)&&(t1<=ix1))) { |
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204 s1 = t1+r; |
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205 if(((int32_t)(t1&sign)==sign)&&(s1&sign)==0) s0 += 1; |
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206 ix0 -= t; |
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207 if (ix1 < t1) ix0 -= 1; |
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208 ix1 -= t1; |
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209 q1 += r; |
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210 } |
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211 ix0 += ix0 + ((ix1&sign)>>31); |
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212 ix1 += ix1; |
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213 r>>=1; |
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214 } |
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215 |
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216 /* use floating add to find out rounding direction */ |
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217 if((ix0|ix1)!=0) { |
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218 z = one-tiny; /* trigger inexact flag */ |
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219 if (z>=one) { |
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220 z = one+tiny; |
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221 if (q1==(u_int32_t)0xffffffff) { q1=0; q += 1;} |
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222 else if (z>one) { |
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223 if (q1==(u_int32_t)0xfffffffe) q+=1; |
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224 q1+=2; |
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225 } else |
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226 q1 += (q1&1); |
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227 } |
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228 } |
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229 ix0 = (q>>1)+0x3fe00000; |
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230 ix1 = q1>>1; |
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231 if ((q&1)==1) ix1 |= sign; |
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232 ix0 += (m <<20); |
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233 INSERT_WORDS(z,ix0,ix1); |
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234 return z; |
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235 } |
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236 |
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237 /* |
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238 Other methods (use floating-point arithmetic) |
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239 ------------- |
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240 (This is a copy of a drafted paper by Prof W. Kahan |
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241 and K.C. Ng, written in May, 1986) |
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242 |
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243 Two algorithms are given here to implement sqrt(x) |
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244 (IEEE double precision arithmetic) in software. |
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245 Both supply sqrt(x) correctly rounded. The first algorithm (in |
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246 Section A) uses newton iterations and involves four divisions. |
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247 The second one uses reciproot iterations to avoid division, but |
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248 requires more multiplications. Both algorithms need the ability |
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249 to chop results of arithmetic operations instead of round them, |
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250 and the INEXACT flag to indicate when an arithmetic operation |
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251 is executed exactly with no roundoff error, all part of the |
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252 standard (IEEE 754-1985). The ability to perform shift, add, |
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253 subtract and logical AND operations upon 32-bit words is needed |
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254 too, though not part of the standard. |
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255 |
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256 A. sqrt(x) by Newton Iteration |
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257 |
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258 (1) Initial approximation |
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259 |
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260 Let x0 and x1 be the leading and the trailing 32-bit words of |
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261 a floating point number x (in IEEE double format) respectively |
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262 |
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263 1 11 52 ...widths |
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264 ------------------------------------------------------ |
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265 x: |s| e | f | |
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266 ------------------------------------------------------ |
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267 msb lsb msb lsb ...order |
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268 |
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269 |
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270 ------------------------ ------------------------ |
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271 x0: |s| e | f1 | x1: | f2 | |
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272 ------------------------ ------------------------ |
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273 |
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274 By performing shifts and subtracts on x0 and x1 (both regarded |
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275 as integers), we obtain an 8-bit approximation of sqrt(x) as |
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276 follows. |
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277 |
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278 k := (x0>>1) + 0x1ff80000; |
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279 y0 := k - T1[31&(k>>15)]. ... y ~ sqrt(x) to 8 bits |
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280 Here k is a 32-bit integer and T1[] is an integer array containing |
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281 correction terms. Now magically the floating value of y (y's |
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282 leading 32-bit word is y0, the value of its trailing word is 0) |
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283 approximates sqrt(x) to almost 8-bit. |
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284 |
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285 Value of T1: |
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286 static int T1[32]= { |
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287 0, 1024, 3062, 5746, 9193, 13348, 18162, 23592, |
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288 29598, 36145, 43202, 50740, 58733, 67158, 75992, 85215, |
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289 83599, 71378, 60428, 50647, 41945, 34246, 27478, 21581, |
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290 16499, 12183, 8588, 5674, 3403, 1742, 661, 130,}; |
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291 |
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292 (2) Iterative refinement |
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293 |
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294 Apply Heron's rule three times to y, we have y approximates |
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295 sqrt(x) to within 1 ulp (Unit in the Last Place): |
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296 |
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297 y := (y+x/y)/2 ... almost 17 sig. bits |
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298 y := (y+x/y)/2 ... almost 35 sig. bits |
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299 y := y-(y-x/y)/2 ... within 1 ulp |
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300 |
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301 |
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302 Remark 1. |
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303 Another way to improve y to within 1 ulp is: |
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304 |
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305 y := (y+x/y) ... almost 17 sig. bits to 2*sqrt(x) |
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306 y := y - 0x00100006 ... almost 18 sig. bits to sqrt(x) |
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307 |
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308 2 |
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309 (x-y )*y |
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310 y := y + 2* ---------- ...within 1 ulp |
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311 2 |
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312 3y + x |
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313 |
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314 |
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315 This formula has one division fewer than the one above; however, |
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316 it requires more multiplications and additions. Also x must be |
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317 scaled in advance to avoid spurious overflow in evaluating the |
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318 expression 3y*y+x. Hence it is not recommended uless division |
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319 is slow. If division is very slow, then one should use the |
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320 reciproot algorithm given in section B. |
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321 |
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322 (3) Final adjustment |
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323 |
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324 By twiddling y's last bit it is possible to force y to be |
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325 correctly rounded according to the prevailing rounding mode |
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326 as follows. Let r and i be copies of the rounding mode and |
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327 inexact flag before entering the square root program. Also we |
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328 use the expression y+-ulp for the next representable floating |
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329 numbers (up and down) of y. Note that y+-ulp = either fixed |
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330 point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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331 mode. |
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332 |
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333 I := FALSE; ... reset INEXACT flag I |
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334 R := RZ; ... set rounding mode to round-toward-zero |
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335 z := x/y; ... chopped quotient, possibly inexact |
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336 If(not I) then { ... if the quotient is exact |
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337 if(z=y) { |
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338 I := i; ... restore inexact flag |
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339 R := r; ... restore rounded mode |
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340 return sqrt(x):=y. |
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341 } else { |
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342 z := z - ulp; ... special rounding |
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343 } |
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344 } |
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345 i := TRUE; ... sqrt(x) is inexact |
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346 If (r=RN) then z=z+ulp ... rounded-to-nearest |
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347 If (r=RP) then { ... round-toward-+inf |
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348 y = y+ulp; z=z+ulp; |
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349 } |
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350 y := y+z; ... chopped sum |
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351 y0:=y0-0x00100000; ... y := y/2 is correctly rounded. |
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352 I := i; ... restore inexact flag |
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353 R := r; ... restore rounded mode |
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354 return sqrt(x):=y. |
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355 |
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356 (4) Special cases |
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357 |
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358 Square root of +inf, +-0, or NaN is itself; |
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359 Square root of a negative number is NaN with invalid signal. |
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360 |
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361 |
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362 B. sqrt(x) by Reciproot Iteration |
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363 |
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364 (1) Initial approximation |
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365 |
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366 Let x0 and x1 be the leading and the trailing 32-bit words of |
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367 a floating point number x (in IEEE double format) respectively |
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368 (see section A). By performing shifs and subtracts on x0 and y0, |
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369 we obtain a 7.8-bit approximation of 1/sqrt(x) as follows. |
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370 |
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371 k := 0x5fe80000 - (x0>>1); |
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372 y0:= k - T2[63&(k>>14)]. ... y ~ 1/sqrt(x) to 7.8 bits |
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373 |
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374 Here k is a 32-bit integer and T2[] is an integer array |
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375 containing correction terms. Now magically the floating |
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376 value of y (y's leading 32-bit word is y0, the value of |
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377 its trailing word y1 is set to zero) approximates 1/sqrt(x) |
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378 to almost 7.8-bit. |
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379 |
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380 Value of T2: |
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381 static int T2[64]= { |
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382 0x1500, 0x2ef8, 0x4d67, 0x6b02, 0x87be, 0xa395, 0xbe7a, 0xd866, |
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383 0xf14a, 0x1091b,0x11fcd,0x13552,0x14999,0x15c98,0x16e34,0x17e5f, |
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384 0x18d03,0x19a01,0x1a545,0x1ae8a,0x1b5c4,0x1bb01,0x1bfde,0x1c28d, |
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385 0x1c2de,0x1c0db,0x1ba73,0x1b11c,0x1a4b5,0x1953d,0x18266,0x16be0, |
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386 0x1683e,0x179d8,0x18a4d,0x19992,0x1a789,0x1b445,0x1bf61,0x1c989, |
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387 0x1d16d,0x1d77b,0x1dddf,0x1e2ad,0x1e5bf,0x1e6e8,0x1e654,0x1e3cd, |
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388 0x1df2a,0x1d635,0x1cb16,0x1be2c,0x1ae4e,0x19bde,0x1868e,0x16e2e, |
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389 0x1527f,0x1334a,0x11051,0xe951, 0xbe01, 0x8e0d, 0x5924, 0x1edd,}; |
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390 |
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391 (2) Iterative refinement |
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392 |
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393 Apply Reciproot iteration three times to y and multiply the |
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394 result by x to get an approximation z that matches sqrt(x) |
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395 to about 1 ulp. To be exact, we will have |
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396 -1ulp < sqrt(x)-z<1.0625ulp. |
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397 |
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398 ... set rounding mode to Round-to-nearest |
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399 y := y*(1.5-0.5*x*y*y) ... almost 15 sig. bits to 1/sqrt(x) |
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400 y := y*((1.5-2^-30)+0.5*x*y*y)... about 29 sig. bits to 1/sqrt(x) |
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401 ... special arrangement for better accuracy |
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402 z := x*y ... 29 bits to sqrt(x), with z*y<1 |
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403 z := z + 0.5*z*(1-z*y) ... about 1 ulp to sqrt(x) |
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404 |
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405 Remark 2. The constant 1.5-2^-30 is chosen to bias the error so that |
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406 (a) the term z*y in the final iteration is always less than 1; |
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407 (b) the error in the final result is biased upward so that |
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408 -1 ulp < sqrt(x) - z < 1.0625 ulp |
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409 instead of |sqrt(x)-z|<1.03125ulp. |
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410 |
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411 (3) Final adjustment |
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412 |
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413 By twiddling y's last bit it is possible to force y to be |
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414 correctly rounded according to the prevailing rounding mode |
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415 as follows. Let r and i be copies of the rounding mode and |
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416 inexact flag before entering the square root program. Also we |
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417 use the expression y+-ulp for the next representable floating |
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418 numbers (up and down) of y. Note that y+-ulp = either fixed |
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419 point y+-1, or multiply y by nextafter(1,+-inf) in chopped |
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420 mode. |
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421 |
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422 R := RZ; ... set rounding mode to round-toward-zero |
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423 switch(r) { |
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424 case RN: ... round-to-nearest |
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425 if(x<= z*(z-ulp)...chopped) z = z - ulp; else |
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426 if(x<= z*(z+ulp)...chopped) z = z; else z = z+ulp; |
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427 break; |
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428 case RZ:case RM: ... round-to-zero or round-to--inf |
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429 R:=RP; ... reset rounding mod to round-to-+inf |
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430 if(x<z*z ... rounded up) z = z - ulp; else |
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431 if(x>=(z+ulp)*(z+ulp) ...rounded up) z = z+ulp; |
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432 break; |
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433 case RP: ... round-to-+inf |
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434 if(x>(z+ulp)*(z+ulp)...chopped) z = z+2*ulp; else |
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435 if(x>z*z ...chopped) z = z+ulp; |
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436 break; |
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437 } |
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438 |
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439 Remark 3. The above comparisons can be done in fixed point. For |
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440 example, to compare x and w=z*z chopped, it suffices to compare |
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441 x1 and w1 (the trailing parts of x and w), regarding them as |
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442 two's complement integers. |
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443 |
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444 ...Is z an exact square root? |
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445 To determine whether z is an exact square root of x, let z1 be the |
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446 trailing part of z, and also let x0 and x1 be the leading and |
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447 trailing parts of x. |
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448 |
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449 If ((z1&0x03ffffff)!=0) ... not exact if trailing 26 bits of z!=0 |
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450 I := 1; ... Raise Inexact flag: z is not exact |
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451 else { |
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452 j := 1 - [(x0>>20)&1] ... j = logb(x) mod 2 |
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453 k := z1 >> 26; ... get z's 25-th and 26-th |
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454 fraction bits |
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455 I := i or (k&j) or ((k&(j+j+1))!=(x1&3)); |
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456 } |
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457 R:= r ... restore rounded mode |
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458 return sqrt(x):=z. |
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459 |
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Sam Lantinga <slouken@libsdl.org>
parents:
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460 If multiplication is cheaper then the foregoing red tape, the |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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461 Inexact flag can be evaluated by |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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462 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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463 I := i; |
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It's now possible to build SDL without any C runtime at all on Windows,
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464 I := (z*z!=x) or I. |
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It's now possible to build SDL without any C runtime at all on Windows,
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465 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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466 Note that z*z can overwrite I; this value must be sensed if it is |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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467 True. |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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468 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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469 Remark 4. If z*z = x exactly, then bit 25 to bit 0 of z1 must be |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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470 zero. |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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471 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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472 -------------------- |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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473 z1: | f2 | |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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474 -------------------- |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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475 bit 31 bit 0 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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476 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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477 Further more, bit 27 and 26 of z1, bit 0 and 1 of x1, and the odd |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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478 or even of logb(x) have the following relations: |
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It's now possible to build SDL without any C runtime at all on Windows,
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parents:
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479 |
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It's now possible to build SDL without any C runtime at all on Windows,
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480 ------------------------------------------------- |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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481 bit 27,26 of z1 bit 1,0 of x1 logb(x) |
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It's now possible to build SDL without any C runtime at all on Windows,
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482 ------------------------------------------------- |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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483 00 00 odd and even |
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484 01 01 even |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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485 10 10 odd |
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It's now possible to build SDL without any C runtime at all on Windows,
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486 10 00 even |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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487 11 01 even |
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It's now possible to build SDL without any C runtime at all on Windows,
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488 ------------------------------------------------- |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
parents:
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489 |
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It's now possible to build SDL without any C runtime at all on Windows,
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490 (4) Special cases (see (4) of Section A). |
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It's now possible to build SDL without any C runtime at all on Windows,
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parents:
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491 |
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It's now possible to build SDL without any C runtime at all on Windows,
Sam Lantinga <slouken@libsdl.org>
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492 */ |
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It's now possible to build SDL without any C runtime at all on Windows,
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493 |