# ldiv Function

Calculates the quotient and remainder of two long integers.

Include

<stdlib.h>

Prototype

ldiv_t ldiv(long numer, long denom);

Arguments

 numer numerator denom denominator

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

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 <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.