Question

Write the definition of a function divide that takes four arguments and returns no value . The first two arguments are of type int . The last two arguments arguments are pointers to int and are set by the function to the quotient and remainder of dividing the first argument by the second argument . The function does not return a value .The function can be used as follows:int numerator=42, denominator=5, quotient, remainder; divide(numerator,denominator,&quotient,&remainder); / quotient is now 8 / / remainder is now 2 /

Answer 1

void divide (int x, int y, int*q, int*r)

{

*q = x / y;

*r = x % y;

}

Step-by-step explanation:

MPL Answer

Answer 2

#include <iostream>

using namespace std;

void divide(int numerator, int denominator, int *quotient, int *remainder)

{

*quotient = (int)(numerator / denominator);

*remainder = numerator % denominator;

}

int main()

{

int num = 42, den = 5, quotient=0, remainder=0;

divide(num, den, &quotient, &remainder);

return 0;

}

Step-by-step explanation:

The exercise is for "Call by pointers". This technique is particularly useful when a variable needs to be changed by a function. In our case, the quotient and the remainder. The '&' is passing by address. Since the function is calling a pointer. We need to pass an address. This way, the function will alter the value at the address.

To sum up, in case we hadn't used pointers here, the quotient and remainder that we set to '0' would have remained zero because the function would've made copies of them, altered the copies and then DELETED the copies. When we pass by pointer, the computer goes inside the memory and changes it at the address. No new copies are made. And the value of the variable is updated.

Thanks! :)