erf(x)和math.h

Question

根据这个站点,错误函数erf(x)来自math.h. 但实际上看看math.h,它不存在,而gcc无法编译以下测试程序,而g ++可以:

#include <math.h>
#include <stdio.h>

int main(int argc, char* argv[]) {
  double x;
  double erfX;
  x = 1.0;
  erfX = erf(x);

  printf("erf(%f) = %f", x, erfX);
}

$ gcc mathHTest.c
/tmp/ccWfNox5.o: In function `main':
mathHTest.c:(.text+0x28): undefined reference to `erf'
collect2: ld returned 1 exit status
$ g++ mathHTest.c

g ++没有g ++带来什么？查看/ usr/include,我能找到的唯一的地方erf(x)是在tgmath.h中,我没有包括.所以g ++必须抓住比gcc更多的标题,但是哪些？

编辑:我没有在libm与gcc链接,因此链接错误.但是,我仍然不明白为什么erf()不在math.h中.它来自哪里？

Answer 1

我有一个类似的问题,需要找到确切的定义,erf所以让我扩展这个.正如Chris Dodd所说,声明了函数,bits/mathcalls.h其中包含了函数maths.h.

bits/mathcalls.h:

...
#if defined __USE_MISC || defined __USE_XOPEN || defined __USE_ISOC99
__BEGIN_NAMESPACE_C99
/* Error and gamma functions.  */
__MATHCALL (erf,, (_Mdouble_));
__MATHCALL (erfc,, (_Mdouble_));
__MATHCALL (lgamma,, (_Mdouble_));
__END_NAMESPACE_C99
#endif
...

宏魔法扩展__MATHCALL (erf,, (_Mdouble_));到

extern double erf (double) throw (); extern double __erf (double) throw ();

实际代码在libm.a或libm.so(gcc -lm)中:

$ nm /usr/lib/libm.a
...
s_erf.o:
00000400 T __erf
00000000 T __erfc
         U __ieee754_exp
00000400 W erf
00000000 W erfc
...

源可以从gnu libc网页获得.关于实际实现的粗略想法,这里有几行来源:

sysdeps/ieee754/dbl-64/s_erf.c:

/* double erf(double x)
 * double erfc(double x)
 *                           x
 *                    2      |\
 *     erf(x)  =  ---------  | exp(-t*t)dt
 *                 sqrt(pi) \|
 *                           0
 *
 *     erfc(x) =  1-erf(x)
 *  Note that
 *              erf(-x) = -erf(x)
 *              erfc(-x) = 2 - erfc(x)
 *
 * Method:
 *      1. For |x| in [0, 0.84375]
 *          erf(x)  = x + x*R(x^2)
 *          erfc(x) = 1 - erf(x)           if x in [-.84375,0.25]
 *                  = 0.5 + ((0.5-x)-x*R)  if x in [0.25,0.84375]
 *         where R = P/Q where P is an odd poly of degree 8 and
 *         Q is an odd poly of degree 10.
 *                                               -57.90
 *                      | R - (erf(x)-x)/x | <= 2
 *
 *
 *         Remark. The formula is derived by noting
 *          erf(x) = (2/sqrt(pi))*(x - x^3/3 + x^5/10 - x^7/42 + ....)
 *         and that
 *          2/sqrt(pi) = 1.128379167095512573896158903121545171688
 *         is close to one. The interval is chosen because the fix
 *         point of erf(x) is near 0.6174 (i.e., erf(x)=x when x is
 *         near 0.6174), and by some experiment, 0.84375 is chosen to
 *         guarantee the error is less than one ulp for erf.
 *
 *      2. For |x| in [0.84375,1.25], let s = |x| - 1, and     
 ...

Answer 2

‘erf’实际上是在bits/mathcalls.h中声明的，它是由math.h#included的。实际的声明被宏魔法严重掩盖，使其对 C 和 C++ 都做正确的事情