6.11.75 fmodf Function
Calculates the remainder of the division between two single-precision floating-point values.
Include
<math.h>
Prototype
float fmodf(float x, float y);
Arguments
x
- a double precision floating-point value
y
- a double precision floating-point value
Return Value
Returns the remainder of x/y
with the same sign as x
and
magnitude less than the magnitude of y
, or returns NaN if
y
is zero.
Remarks
If y
is zero, a domain error occurs, and errno
will be
set to EDOM
.
Example
See the notes at the beginning of this chapter or section for
information on using printf()
or scanf()
(and other functions reading and writing the stdin
or
stdout
streams) in the example code.
#include <math.h>
#include <stdio.h>
#include <errno.h>
int main(void)
{
float x,y,z;
errno = 0;
x = 7.0F;
y = 3.0F;
z = fmodf(x, y);
if(errno)
perror("Error");
printf("For fmodf(%f, %f) the remainder is"
" %f\n\n", x, y, z);
errno = 0;
x = -5.0F;
y = 3.0F;
z = fmodf(x, y);
if(errno)
perror("Error");
printf("For fmodf(%f, %f) the remainder is %f\n", x, y, z);
errno = 0;
x = 5.0F;
y = -3.0F;
z = fmodf(x, y);
if(errno)
perror("Error");
printf("For fmodf(%f, %f) the remainder is %f\n", x, y, z);
errno = 0;
x = 5.0F;
y = -5.0F;
z = fmodf (x, y);
if(errno)
perror("Error");
printf("For fmodf (%f, %f) the remainder is %f\n", x, y, z);
errno = 0;
x = 7.0F;
y = 0.0F;
z = fmodf(x, y);
if(errno)
perror("Error");
printf("For fmodf(%f, %f) the remainder is %f\n", x, y, z);
errno = 0;
x = 7.0F;
y = 7.0F;
z = fmodf(x, y);
if(errno)
perror("Error");
printf("For fmodf(%f, %f) the remainder is %f\n", x, y, z);
}
Example Output
For fmodf (7.000000, 3.000000) the remainder is 1.000000
For fmodf (-5.000000, 3.000000) the remainder is -2.000000
For fmodf (5.000000, -3.000000) the remainder is 2.000000
For fmodf (5.000000, -5.000000) the remainder is 0.000000
Error: domain error
For fmodf (7.000000, 0.000000) the remainder is nan
For fmodf (7.000000, 7.000000) the remainder is 0.000000