e_log2.c 3.8 KB

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  1. /* Adapted for log2 by Ulrich Drepper <drepper@cygnus.com>. */
  2. /*
  3. * ====================================================
  4. * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
  5. *
  6. * Developed at SunPro, a Sun Microsystems, Inc. business.
  7. * Permission to use, copy, modify, and distribute this
  8. * software is freely granted, provided that this notice
  9. * is preserved.
  10. * ====================================================
  11. */
  12. /* __ieee754_log2(x)
  13. * Return the logarithm to base 2 of x
  14. *
  15. * Method :
  16. * 1. Argument Reduction: find k and f such that
  17. * x = 2^k * (1+f),
  18. * where sqrt(2)/2 < 1+f < sqrt(2) .
  19. *
  20. * 2. Approximation of log(1+f).
  21. * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s)
  22. * = 2s + 2/3 s**3 + 2/5 s**5 + .....,
  23. * = 2s + s*R
  24. * We use a special Reme algorithm on [0,0.1716] to generate
  25. * a polynomial of degree 14 to approximate R The maximum error
  26. * of this polynomial approximation is bounded by 2**-58.45. In
  27. * other words,
  28. * 2 4 6 8 10 12 14
  29. * R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s +Lg6*s +Lg7*s
  30. * (the values of Lg1 to Lg7 are listed in the program)
  31. * and
  32. * | 2 14 | -58.45
  33. * | Lg1*s +...+Lg7*s - R(z) | <= 2
  34. * | |
  35. * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2.
  36. * In order to guarantee error in log below 1ulp, we compute log
  37. * by
  38. * log(1+f) = f - s*(f - R) (if f is not too large)
  39. * log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy)
  40. *
  41. * 3. Finally, log(x) = k + log(1+f).
  42. * = k+(f-(hfsq-(s*(hfsq+R))))
  43. *
  44. * Special cases:
  45. * log2(x) is NaN with signal if x < 0 (including -INF) ;
  46. * log2(+INF) is +INF; log(0) is -INF with signal;
  47. * log2(NaN) is that NaN with no signal.
  48. *
  49. * Constants:
  50. * The hexadecimal values are the intended ones for the following
  51. * constants. The decimal values may be used, provided that the
  52. * compiler will convert from decimal to binary accurately enough
  53. * to produce the hexadecimal values shown.
  54. */
  55. #include "math.h"
  56. #include "math_private.h"
  57. #ifdef __STDC__
  58. static const double
  59. #else
  60. static double
  61. #endif
  62. ln2 = 0.69314718055994530942,
  63. two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */
  64. Lg1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */
  65. Lg2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */
  66. Lg3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */
  67. Lg4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */
  68. Lg5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */
  69. Lg6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */
  70. Lg7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */
  71. #ifdef __STDC__
  72. static const double zero = 0.0;
  73. #else
  74. static double zero = 0.0;
  75. #endif
  76. #ifdef __STDC__
  77. double __ieee754_log2(double x)
  78. #else
  79. double __ieee754_log2(x)
  80. double x;
  81. #endif
  82. {
  83. double hfsq,f,s,z,R,w,t1,t2,dk;
  84. int32_t k,hx,i,j;
  85. u_int32_t lx;
  86. EXTRACT_WORDS(hx,lx,x);
  87. k=0;
  88. if (hx < 0x00100000) { /* x < 2**-1022 */
  89. if (((hx&0x7fffffff)|lx)==0)
  90. return -two54/(x-x); /* log(+-0)=-inf */
  91. if (hx<0) return (x-x)/(x-x); /* log(-#) = NaN */
  92. k -= 54; x *= two54; /* subnormal number, scale up x */
  93. GET_HIGH_WORD(hx,x);
  94. }
  95. if (hx >= 0x7ff00000) return x+x;
  96. k += (hx>>20)-1023;
  97. hx &= 0x000fffff;
  98. i = (hx+0x95f64)&0x100000;
  99. SET_HIGH_WORD(x,hx|(i^0x3ff00000)); /* normalize x or x/2 */
  100. k += (i>>20);
  101. dk = (double) k;
  102. f = x-1.0;
  103. if((0x000fffff&(2+hx))<3) { /* |f| < 2**-20 */
  104. if(f==zero) return dk;
  105. R = f*f*(0.5-0.33333333333333333*f);
  106. return dk-(R-f)/ln2;
  107. }
  108. s = f/(2.0+f);
  109. z = s*s;
  110. i = hx-0x6147a;
  111. w = z*z;
  112. j = 0x6b851-hx;
  113. t1= w*(Lg2+w*(Lg4+w*Lg6));
  114. t2= z*(Lg1+w*(Lg3+w*(Lg5+w*Lg7)));
  115. i |= j;
  116. R = t2+t1;
  117. if(i>0) {
  118. hfsq=0.5*f*f;
  119. return dk-((hfsq-(s*(hfsq+R)))-f)/ln2;
  120. } else {
  121. return dk-((s*(f-R))-f)/ln2;
  122. }
  123. }