In C++, does Function declaration play a role in the way function parameters are passed - either by reference or by Value

Question

I understand without &, the values are passed by value. Consider the following code snippet:

void swap(int x, int y)
{
    int tmp = x;
    x = y; 
    y = tmp;
}


int main()
{
    int a=10, b=20;
    swap(a,b);
    cout << a << " " << b << endl;
    return 0;
}

The output will be : 10 20

However if the swap function is placed after the main function and there is no function declaration before the main. That is...


int main()
{
    int a=10, b=20;
    swap(a,b);
    cout << a << " " << b << endl;
    return 0;
}

void swap(int x, int y)
{
    int tmp = x;
    x = y; 
    y = tmp;
}

In this case the output is going to be : 20 10.

Shouldn't the o/p also be 10 20 in the second case. Can someone please help me to understand as to ow is pass by reference happening here?

Get rid of `using namespapce std;`. 99.9% sure you are accidently pulling in `std::swap`: https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice — NathanOliver, Jun 15 '21 at 12:05
@adnan_e anything is possible when looking at incomplete snippets of code. I can imagine uncountable other ways to add missing code to reproduce the reported output. — 463035818_is_not_an_ai, Jun 15 '21 at 12:08
Please read about [mcve] and post one. The important part is in code you did not post — 463035818_is_not_an_ai, Jun 15 '21 at 12:09
Removing the using namespace std did the trick. Thanks @NathanOliver. — Vamsi Krishna, Jun 15 '21 at 12:14
Sure @463035818_is_not_a_number, will make sure to add the complete code snippet next time. — Vamsi Krishna, Jun 15 '21 at 12:15

In C++, does Function declaration play a role in the way function parameters are passed - either by reference or by Value

0 Answers0