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在头文件<math.h>中定义



float       scalbnf( float arg, int exp );

(1)

(since C99)

double      scalbn( double arg, int exp );

(2)

(since C99)

long double scalbnl( long double arg, int exp );

(3)

(since C99)

Defined in header <tgmath.h>



#define scalbn( arg, exp )

(4)

(since C99)

Defined in header <math.h>



float       scalblnf( float arg, long exp );

(5)

(since C99)

double      scalbln( double arg, long exp );

(6)

(since C99)

long double scalblnl( long double arg, long exp );

(7)

(since C99)

Defined in header <tgmath.h>



#define scalbln( arg, exp )

(8)

(since C99)

1-3,5-7)将浮点值arg乘以FLT_RADIX功率exp

如果实现支持IEEE浮点运算(IEC 60559),对于roundroundfroundl功能:

  • 当前的舍入模式不起作用。

  • 如果arg是±∞,则返回,未修改

  • 如果arg为±0,则返回,未修改

  • 如果arg是NaN,则返回NaN

对于lroundllround功能家庭:

  • FE_INEXACT 从未被提出

  • 当前的舍入模式不起作用。

  • 如果arg是±∞,FE_INVALID则引发并返回实现定义的值

  • 如果舍入的结果超出返回类型的范围,FE_INVALID则会引发并返回实现定义的值

  • 如果arg是NaN,FE_INVALID则引发并返回实现定义的值

笔记

FE_INEXACT可能是(但不要求)round在舍入非整数有限值时引发。

最大的可表示浮点值是所有标准浮点格式中的精确整数,因此round不会自行溢出; 但是intmax_t,当存储在整数变量中时,结果可能会溢出任何整数类型(包括)。

POSIX规定是,所有病例lroundllround提高FE_INEXACT是域错误。

该行为的double版本round如同执行如下:

按照math_errhandling中的指定报告错误。

#include <math.h>#include <fenv.h>#pragma STDC FENV_ACCESS ON
double round(double x){
    fenv_t save_env;    feholdexcept(&save_env);
    double result = rint(x);    if (fetestexcept(FE_INEXACT)) {        fesetround(FE_TOWARDZERO);
        result = rint(copysign(0.5 + fabs(x), x));    }    feupdateenv(&save_env);    return result;}
  • 除非发生范围错误,否则FE_INEXACT不会引发(结果是确切的)

  • 除非发生范围错误,否则当前舍入模式将被忽略

  • 如果arg为±0,则返回,未修改

  • 如果arg是±∞,则返回,未修改

  • 如果exp为0,则arg返回,未修改

  • 如果arg是NaN,则返回NaN

二进制系统(其中,FLT_RADIX2),scalbn相当于ldexp

#include <stdio.h>#include <math.h>#include <fenv.h>#include <limits.h>
 #pragma STDC FENV_ACCESS ON
 
int main(void){    // round    printf("round(+2.3) = %+.1f  ", round(2.3));    printf("round(+2.5) = %+.1f  ", round(2.5));    printf("round(+2.7) = %+.1f\n", round(2.7));    printf("round(-2.3) = %+.1f  ", round(-2.3));    printf("round(-2.5) = %+.1f  ", round(-2.5));    printf("round(-2.7) = %+.1f\n", round(-2.7)); 
    printf("round(-0.0) = %+.1f\n", round(-0.0));    printf("round(-Inf) = %+f\n",   round(-INFINITY)); 
    // lround    printf("lround(+2.3) = %ld  ", lround(2.3));    printf("lround(+2.5) = %ld  ", lround(2.5));    printf("lround(+2.7) = %ld\n", lround(2.7));    printf("lround(-2.3) = %ld  ", lround(-2.3));    printf("lround(-2.5) = %ld  ", lround(-2.5));    printf("lround(-2.7) = %ld\n", lround(-2.7)); 
    printf("lround(-0.0) = %ld\n", lround(-0.0));    printf("lround(-Inf) = %ld\n", lround(-INFINITY)); // FE_INVALID raised 
    // error handling    feclearexcept(FE_ALL_EXCEPT);    printf("lround(LONG_MAX+1.5) = %ld\n", lround(LONG_MAX+1.5));    if(fetestexcept(FE_INVALID)) puts("    FE_INVALID was raised");}

scalbln提供该函数是因为从最小正浮点值缩放到最大有限值所需的因子可能大于标准保证的32767 INT_MAX。特别是对于80位long double,因子是32828。

可能的输出:

#include <stdio.h>#include <math.h>#include <float.h>#include <errno.h>#include <fenv.h>#pragma STDC FENV_ACCESS ON
int main(void){    printf("scalbn(7, -4) = %f\n", scalbn(7, -4));    printf("scalbn(1, -1074) = %g (minimum positive subnormal double)\n",            scalbn(1, -1074));    printf("scalbn(nextafter(1,0), 1024) = %g (largest finite double)\n",            scalbn(nextafter(1,0), 1024));    // special values    printf("scalbn(-0, 10) = %f\n", scalbn(-0.0, 10));    printf("scalbn(-Inf, -1) = %f\n", scalbn(-INFINITY, -1));    //error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);    printf("scalbn(1, 1024) = %f\n", scalbn(1, 1024));    if(errno == ERANGE) perror("    errno == ERANGE");    if(fetestexcept(FE_OVERFLOW)) puts("    FE_OVERFLOW raised");}
  • C11标准(ISO / IEC 9899:2011):

    • 7.12.9.6轮次函数(p:253)

    • 7.12.9.7地下和地下功能(p:253)

    • 7.25类型通用数学<tgmath.h>(p:373-375)

    • F.10.6.6循环函数(p:527)

    • F.10.6.7基础和基本功能(p:528)

  • C99标准(ISO / IEC 9899:1999):

    • 7.12.9.6轮次函数(p:233)

    • 7.12.9.7理论和实践(p:234)

    • 7.22类型通用数学<tgmath.h>(p:335-337)

    • F.9.6.6循环函数(p:464)

    • F.9.6.7地下和地下功能(p:464)

scalbn(7, -4) = 0.437500scalbn(1, -1074) = 4.94066e-324 (minimum positive subnormal double)scalbn(nextafter(1,0), 1024) = 1.79769e+308 (largest finite double)scalbn(-0, 10) = -0.000000scalbn(-Inf, -1) = -infscalbn(1, 1024) = inf
    errno == ERANGE: Numerical result out of range
    FE_OVERFLOW raised

参考

  • C11标准(ISO / IEC 9899:2011):

    • 7.12.6.13 scalbn函数(p:247)

    • 7.25类型通用数学<tgmath.h>(p:373-375)

    • F.10.3.13 scalbn函数(p:523)

  • C99标准(ISO / IEC 9899:1999):

    • 7.12.6.13 scalbn函数(p:228)

    • 7.22类型通用数学<tgmath.h>(p:335-337)

    • F.9.3.13 scalbn函数(p:460)

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