The program below reads two integers. Then a function with the following prototype is called: void div_rem(int a, int b, int *divides, int *remains); This function is passed the two integers. It divides them (using integer division), and writes the answer over wherever “divides” points. Then it finds the remainder and writes it into where “remains” points. Thus for 20 and 3, 20 divided by 3 is 6, remainder 2. Implement the div_rem function. #include void div_rem(int a, int b, int *divides, int *remains); int main(void)
Question 3
The program below reads two integers. Then a function with the following
prototype is called:
void div_rem(int a, int b, int *divides, int *remains);
This function is passed the two integers. It divides them (using integer division),
and writes the answer over wherever “divides” points. Then it finds the
remainder and writes it into where “remains” points. Thus for 20 and 3, 20
divided by 3 is 6, remainder 2. Implement the div_rem function.
#include <stdio.h>
void div_rem(int a, int b, int *divides, int *remains);
int main(void)
{
int a, b;
int div = 0;
int rem = 0;
printf("enter two integers ");
scanf("%i %i", &a, &b);
div_rem(a, b, &div, &rem);
printf("%i divided by %i = %i "
"remainder %i\n", a, b, div, rem);
return 0;
}
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