CODEWITHSUNDEEP
c++pointerspass-by-reference
namith
namith

Reputation: 548

std::cout not printing to console

#include <iostream>

void swap(int &pi, int &pj){

    std::cout << "In function swap: " <<  &pi << " " << &pj << "\n";

        int temp = pi;
        pi = pj;
        pj = temp;
}

int main(){

        int i = 10, j = 20;
        int *pi = &i, *pj = &j;
        swap(pi, pj);

        std::cout << *pi << " " << *pj;
        return 0;
}

The above program does not give any compilation error. (Though to swap function in not POINTER TO REFERENCE type) and gives the proper output. But whatever i am trying to print inside "swap" function is not printed to console. Can anybody explain me why?

Upvotes: 1

Views: 2148

Answers (3)

sebm
sebm

Reputation: 1

If you can compile your program, you don't show us all of your code. This code snippet you posted doesn't compile, because you're trying to pass to objects of type int* to your function swapm which expects a reference to an int.

If your code compiles, I suspect you to include a 'using namespace std;' anywhere in your original code.

Upvotes: 0

John Corbett
John Corbett

Reputation: 1615

you're trying to get the address of a pointer that you're passing by reference... I don't think you quite understand what you're doing. the pass by reference feature in C++ allows you to pass a reference, so you don't need a pointer. Your code should look like this. 

#include <iostream>

void swap(int &i, int &j){

    std::cout << "In function swap: " <<  i << " " << j << "\n";

        int temp = i;
        i = j;
        j = temp;
}

int main(){

        int i = 10, j = 20;
        swap(i, j);

        std::cout << i << " " << j;
        return 0;
}

Upvotes: -1

Paul R
Paul R

Reputation: 213059

Looks like you're probably using std::swap to swap two pointers instead of calling your own swap routine. I suspect you have a using namespace std; somewhere that you are not showing us ? Try changing the name of your swap routine to e.g. my_swap and then see if calling my_swap works (it should fail with a compilation error).

Upvotes: 9

Related Questions