Calculates the quotient and remainder of two long integers.
Include
<stdlib.h>
Prototype
ldiv_t ldiv(long numer, long denom);
Arguments
numer
denom
Return Value
Returns the quotient and the remainder.
Remarks
The returned quotient will have the same sign as the numerator divided by the
denominator. The sign for the remainder will be such that the quotient times the
denominator plus the remainder will equal the numerator (quot * denom + rem =
numer
). If the denominator is zero, the behavior is undefined.
Example
#include <stdlib.h>
#include <stdio.h>
int main(void)
{
long x,y;
ldiv_t z;
x = 7;
y = 3;
printf("For ldiv(%ld, %ld)\n", x, y);
z = ldiv(x, y);
printf("The quotient is %ld and the "
"remainder is %ld\n\n", z.quot, z.rem);
x = 7;
y = -3;
printf("For ldiv(%ld, %ld)\n", x, y);
z = ldiv(x, y);
printf("The quotient is %ld and the "
"remainder is %ld\n\n", z.quot, z.rem);
x = -5;
y = 3;
printf("For ldiv(%ld, %ld)\n", x, y);
z = ldiv(x, y);
printf("The quotient is %ld and the "
"remainder is %ld\n\n", z.quot, z.rem);
x = 7;
y = 7;
printf("For ldiv(%ld, %ld)\n", x, y);
z = ldiv(x, y);
printf("The quotient is %ld and the "
"remainder is %ld\n\n", z.quot, z.rem);
x = 7;
y = 0;
printf("For ldiv(%ld, %ld)\n", x, y);
z = ldiv(x, y);
printf("The quotient is %ld and the "
"remainder is %ld\n\n",
z.quot, z.rem);
}
Example Output
For ldiv(7, 3)
The quotient is 2 and the remainder is 1
For ldiv(7, -3)
The quotient is -2 and the remainder is 1
For ldiv(-5, 3)
The quotient is -1 and the remainder is -2
For ldiv(7, 7)
The quotient is 1 and the remainder is 0
For ldiv(7, 0)
The quotient is -1 and the remainder is 7
Example Explanation
In the last example (ldiv(7,0)
) the denominator is zero, the behavior is
undefined.