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📄 s_lrint.c

📁 Newlib 嵌入式 C库 标准实现代码
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/* @(#)s_lrint.c 5.1 93/09/24 *//* * ==================================================== * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. * * Developed at SunPro, a Sun Microsystems, Inc. business. * Permission to use, copy, modify, and distribute this * software is freely granted, provided that this notice  * is preserved. * ==================================================== *//* * lrint(x) * Return x rounded to integral value according to the prevailing * rounding mode. * Method: *	Using floating addition. * Exception: *	Inexact flag raised if x not equal to lrint(x). */#include "fdlibm.h"#ifndef _DOUBLE_IS_32BITS#ifdef __STDC__static const double#elsestatic double #endif/* Adding a double, x, to 2^52 will cause the result to be rounded based on   the fractional part of x, according to the implementation's current rounding   mode.  2^52 is the smallest double that can be represented using all 52 significant   digits. */TWO52[2]={  4.50359962737049600000e+15, /* 0x43300000, 0x00000000 */ -4.50359962737049600000e+15, /* 0xC3300000, 0x00000000 */};#ifdef __STDC__	long int lrint(double x)#else	long int lrint(x)	double x;#endif{  __int32_t i0,j0,sx;  __uint32_t i1;  double t;  volatile double w;  long int result;    EXTRACT_WORDS(i0,i1,x);  /* Extract sign bit. */  sx = (i0>>31)&1;  /* Extract exponent field. */  j0 = ((i0 & 0x7ff00000) >> 20) - 1023;    if(j0 < 20)    {      if(j0 < -1)        return 0;      else        {          w = TWO52[sx] + x;          t = w - TWO52[sx];          GET_HIGH_WORD(i0, t);          /* Detect the all-zeros representation of plus and             minus zero, which fails the calculation below. */          if ((i0 & ~(1 << 31)) == 0)              return 0;          j0 = ((i0 & 0x7ff00000) >> 20) - 1023;          i0 &= 0x000fffff;          i0 |= 0x00100000;          result = i0 >> (20 - j0);        }    }  else if (j0 < (int)(8 * sizeof (long int)) - 1)    {      if (j0 >= 52)        result = ((long int) ((i0 & 0x000fffff) | 0x0010000) << (j0 - 20)) |                    (i1 << (j0 - 52));      else        {          w = TWO52[sx] + x;          t = w - TWO52[sx];          EXTRACT_WORDS (i0, i1, t);          j0 = ((i0 & 0x7ff00000) >> 20) - 1023;          i0 &= 0x000fffff;          i0 |= 0x00100000;          result = ((long int) i0 << (j0 - 20)) | (i1 >> (52 - j0));        }    }  else    {      return (long int) x;    }    return sx ? -result : result;}#endif /* _DOUBLE_IS_32BITS */

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