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Home/ Questions/Q 9177423
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T17:19:25+00:00 2026-06-17T17:19:25+00:00

#include <iostream> using namespace std; void swap(int *a, int *b) { *a = *a^*b;

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#include <iostream>

using namespace std;

void swap(int *a, int *b) {
    *a = *a^*b;
    *b = *a^*b;
    *a = *a^*b;
}

int main()
{
    int array[]={1,9,2,8,3,7};
    for(int i=0; i<6; i++)
        cout<<array[i];
    cout<<endl;
    swap(array[1], array[4]);
    for(int i=0; i<6;i++)
        cout<<array[i];
    cout<<endl;
    return 0;
}

above is a test sample. I find if I use swap(array[1], array[4]);, it also swaps the values of two positions in array. But this confuses me, because the function swap() needs two pointers, not two integer values.

Thanks for your help:)

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  1. Editorial Team
    Editorial Team
    2026-06-17T17:19:27+00:00Added an answer on June 17, 2026 at 5:19 pm
    using namespace std;  
    

    This is your culprit. When you import the std:: namespace, you get every identifier declared in that namespace, potentially including std::swap.

    Thus you are invoking std::swap<int>(int&,int&) (from the standard library) and not ::swap(int*,int*) (from your program.)

    The moral of the story: never say using namespace std;. It’s just too big.

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