CODEWITHSUNDEEP
c++
Hyunsu
Hyunsu

Reputation: 33

how do I make multiple integers pass multiple functions?

In this code I am trying to output the sum and difference of two inputs. When I run this program, the sum outputs fine, but instead of the difference, the first input outputs. How can I make this program output the difference as well as the sum?

void Modify(int *x, int *y) {
    *x = *x + *y;
    
    if (*x >= *y)
        *y = *x - *y;
    if (*x < *y)
        *y = *y - *x;
                
}

int main() {
    
    int x, y;
    
    int *ptr_x = &x;
    int *ptr_y = &y;
    
    cin >> x;
    cin >> y;
    
    Modify(ptr_x, ptr_y);
    
    cout << x << endl;
    cout << y;
                 
    return 0;
}

Upvotes: 1

Views: 278

Answers (3)

Picaud Vincent
Picaud Vincent

Reputation: 10982

void Modify(int *x, int *y) {
    *x = *x + *y;
    
    if (*x >= *y)
        *y = *x - *y;
    if (*x < *y)
        *y = *y - *x;
                
}

Your first line *x = *x + *y; overwrites *x value with *x + *y, by consequence the next statements are misleading

    if (*x >= *y)
        *y = *x - *y;
    if (*x < *y)
        *y = *y - *x;

and are not performing what you expect.

As a general advice, keep your code simple and clear about your intends:

#include <cmath>

void Modify(int *x, int *y) {

    using std::abs;

    int sum      = *x + *y;
    int abs_diff = abs(*x - *y);
    
    *x = sum;
    *y = abs_diff;
}

Upvotes: 5

foragerDev
foragerDev

Reputation: 1409

This is not printing difference because you update the value of x with the sum of both at *x = *x + *y as @user4581301 suggested in the comment. You can do something like this.

 
void Modify(int *x, int *y) {
    int* temp = new int(*x);
    *x = *temp + *y;
    
    if (*temp >= *y)
        *y = *temp - *y;
    if (*x < *y)
        *y = *y - *temp;
    delete temp;
}

What this will do is you will create an new integer pointertemp will store the value of x and now you can change x but on other places where x was needed we will use temp instead of x. And at the end we are deleting the temp.

Upvotes: 1

Eduardo Pascual Aseff
Eduardo Pascual Aseff

Reputation: 1166

It would be nice if you explain why you cannot do something simple as std::cout << x + y << endl << x - y; without calling any function.

But if you want to do it that way, without extra variables, one option is this:

void Modify(int *x, int *y) {
    *x = *x + *y;      ///x = original x + original y
    *y = *x - 2*(*y);  ///y = (original x + original y) - 2*(original y)        
}

Note: Be carefull with overflows

Edit: If you want absolute value of the difference, use instead *y = abs(*x - 2*(*y));

Upvotes: 2

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