单、双精度的浮点数为什么不能进行取余运算？

[a1988lourong]

单、双精度的浮点数为什么不能进行取余运算？

[LinuxBirdMan]

问C语言之父去，他在那里等你~~

[飞天御剑流]

%是用于整数的，浮点数用fmod函数。

[quwei197874]

[Quote=引用 2 楼 supermegaboy 的回复:]
%是用于整数的，浮点数用fmod函数。
[/Quote]++

[赵4老师]

fmod
Calculates the floating-point remainder.

double fmod( double x, double y );

Function Required Header Compatibility
fmod <math.h> ANSI, Win 95, Win NT


For additional compatibility information, see Compatibility in the Introduction.

Libraries

LIBC.LIB Single thread static library, retail version
LIBCMT.LIB Multithread static library, retail version
MSVCRT.LIB Import library for MSVCRT.DLL, retail version


Return Value

fmod returns the floating-point remainder of x / y. If the value of y is 0.0, fmod returns a quiet NaN. For information about representation of a quiet NaN by the printf family, see printf.

Parameters

x, y

Floating-point values

Remarks

The fmod function calculates the floating-point remainder f of x / y such that x = i * y + f, where i is an integer, f has the same sign as x, and the absolute value of f is less than the absolute value of y.

Example

/* FMOD.C: This program displays a
* floating-point remainder.
*/

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

void main( void )
{
double w = -10.0, x = 3.0, y = 0.0, z;

z = fmod( x, y );
printf( "The remainder of %.2f / %.2f is %f\n", w, x, z );
printf( "The remainder of %.2f / %.2f is %f\n", x, y, z );

}


Output

The remainder of -10.00 / 3.00 is -1.000000


Floating-Point Support Routines

See Also ceil, fabs, floor

[a1988lourong]

谢谢各位的帮忙

[追求执着]

这个我也不知道，学习楼上！

[luciferisnotsatan]

[Quote=引用 2 楼 supermegaboy 的回复:]

%是用于整数的，浮点数用fmod函数。
[/Quote]
++

[hulongchuan]

因为带小数点的，永远也求不出他的余数是多少！自然也就不能取模。