问题描述
此代码，使用visual studio .NET 2003编译，
double a = 95.022，b = 0.01;
printf（"％lf - 楼层（％lf /％lf）*％lf =％。17lf \ n，a，a，b，b，a -
楼层（a / b）* b）;
a = 95.021，b = 0.01;
printf（"％lf - floor（％lf /％lf）*％lf =％。17lf \ n"，a， a，b，b，a -
楼（a / b）* b）;
a = 95.020，b = 0.01;
printf （"％lf - floor（％lf /％lf）*％lf =％。17lf \ n"，a，a，b，b，a -
楼（a / b） * b）;
a = 95.022，b = 0.01;
printf（" fmod（％lf，％lf）=％。17lf \ n" ;，a，b，fmod（a，b））;
a = 95.021，b = 0.01;
printf（" fmod（％lf，％lf）= ％。17lf \ n"，a，b，fmod（a，b））;
a = 95.020，b = 0.01;
printf（" fmod（％） lf，％lf）=％。17lf \ n"，a，b，fmod（a，b））;
生成此输出：
95.022000 - floor（95.022000 / 0.010000）* 0.010000 =
0.00200000000000955
95.021000 - 楼层（95.021000 / 0.010000）* 0.010000 =
0.00100000000000477
95.020000 - 楼层（95.020000 / 0.010000）* 0.010000 =
0.00000000000000000
fmod（95.022000,0.010000）= 0.00200000000000359
fmod（95.021000,0.010000）= 0.00099999999999882
fmod（95.020000,0.010000）= 0.00999999999999404
一切都有意义，除了最后一行。为什么fmod返回
0.01而不是0.0？
This code, compiled with visual studio .NET 2003,
double a = 95.022, b = 0.01;
printf ("%lf - floor(%lf / %lf) * %lf = %.17lf\n", a, a, b, b, a -
floor(a / b) * b);
a = 95.021, b = 0.01;
printf ("%lf - floor(%lf / %lf) * %lf = %.17lf\n", a, a, b, b, a -
floor(a / b) * b);
a = 95.020, b = 0.01;
printf ("%lf - floor(%lf / %lf) * %lf = %.17lf\n", a, a, b, b, a -
floor(a / b) * b);
a = 95.022, b = 0.01;
printf ("fmod(%lf, %lf) = %.17lf\n", a, b, fmod(a, b));
a = 95.021, b = 0.01;
printf ("fmod(%lf, %lf) = %.17lf\n", a, b, fmod(a, b));
a = 95.020, b = 0.01;
printf ("fmod(%lf, %lf) = %.17lf\n", a, b, fmod(a, b));
generates this output:
95.022000 - floor(95.022000 / 0.010000) * 0.010000 =
0.00200000000000955
95.021000 - floor(95.021000 / 0.010000) * 0.010000 =
0.00100000000000477
95.020000 - floor(95.020000 / 0.010000) * 0.010000 =
0.00000000000000000
fmod(95.022000, 0.010000) = 0.00200000000000359
fmod(95.021000, 0.010000) = 0.00099999999999882
fmod(95.020000, 0.010000) = 0.00999999999999404
everything makes sense, except for the last line. why does fmod return
0.01 instead of 0.0?
推荐答案
我不知道。
我得到的结果与我相似自制fmod。
/ * BEGIN new.c输出* /
fs_fmod（95.022000,0.010000）是0.002000
fs_fmod（95.021000,0.010000）是0.001000
fs_fmod（95.020000,0.010000）是0.010000
/ * END new.c输出* /
/ * BEGIN new.c * /
#include< stdio.h>
#include< float .h>
double fs_fmod（double x，double y）;
int main（无效）
{
puts（" / * BEGIN new.c output * / \ n"）;
printf（" fs_fmod（95.022000,0.010000）is％f \\ \\ n"
，fs_fmod（95.022000,0.010000））;
printf（" fs_fmod（95.021000,0.010000）is％f\ n"
，fs_fmod（95.021000,0.010000））;
printf（" fs_fmod（95.020000,0.010000）是％f \\ n n"
，fs_fmod（95.020000,0.010000））;
puts（" \ n / * END new.c output * /"）;
返回0;
}
double fs_fmod（double x，double y）
{
加倍a，b;
const double c = x;
if（0> c）{
x = -x;
}
if（0> y）{
y = - y;
}
if（y！= 0&& DBL_MAX> = y&& DBL_MAX> = x）{
while（x> = y）{
a = x / 2;
b = y;
while（a> = b）{
b * = 2;
}
x - = b;
}
} else {
x = 0;
}
返回0> C ？ -x：x;
}
/ * END new.c * /
-
pete
I don''t know.
I get similar results with my homemade fmod.
/* BEGIN new.c output */
fs_fmod(95.022000, 0.010000) is 0.002000
fs_fmod(95.021000, 0.010000) is 0.001000
fs_fmod(95.020000, 0.010000) is 0.010000
/* END new.c output */
/* BEGIN new.c */
#include <stdio.h>
#include <float.h>
double fs_fmod(double x, double y);
int main(void)
{
puts("/* BEGIN new.c output */\n");
printf("fs_fmod(95.022000, 0.010000) is %f\n"
, fs_fmod(95.022000, 0.010000));
printf("fs_fmod(95.021000, 0.010000) is %f\n"
, fs_fmod(95.021000, 0.010000));
printf("fs_fmod(95.020000, 0.010000) is %f\n"
, fs_fmod(95.020000, 0.010000));
puts("\n/* END new.c output */");
return 0;
}
double fs_fmod(double x, double y)
{
double a, b;
const double c = x;
if (0 > c) {
x = -x;
}
if (0 > y) {
y = -y;
}
if (y != 0 && DBL_MAX >= y && DBL_MAX >= x) {
while (x >= y) {
a = x / 2;
b = y;
while (a >= b) {
b *= 2;
}
x -= b;
}
} else {
x = 0;
}
return 0 > c ? -x : x;
}
/* END new.c */
--
pete
因为你的程序中没有任何文字可以表示
完全在二进制浮点数。
请参阅comp.lang.c常见问题解答的第14节，< http://www.c-faq.com/> ;.
-
Keith Thompson（The_Other_Keith）< http://www.ghoti.net/~kst>
圣地亚哥超级计算机中心< *> < http://users.sdsc.edu/~kst>
我们必须做点什么。这是事情。因此，我们必须这样做。
Because none of the literals in your program can be represented
exactly in binary floating-point.
See section 14 of the comp.lang.c FAQ, <http://www.c-faq.com/>.
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
这篇关于FMOD的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！