[solved] why swap function is working well without prototyping? (C++)

Question

Because I am using swap() function inside main() function so swap() function should be defined before the main() function OR it should be prototyped before main() function. But here in this program

https://code.sololearn.com/cbD2NDDWru9J/?ref=app

I did neither of them but still it is working without errors or warnings.
On the other hand, if I try to do this with diff() function it gives error.

Can anyone explain why this is not happening with swap() function?

Answer 1

That is because you are not calling your own swap() function, but the one from the standard library, which is indirectly included via <iostream> in this case:

https://en.cppreference.com/w/cpp/algorithm/swap

You can verify this by trying to print anything inside your own swap() function. Also, since you are using pointers instead of references, the correct call to your own function would be this anyway:

swap( &a, &b );

If you tried that without adding the prototype, you would then get an error as expected, because std::swap() is not overladed for addresses.

Answer 2

You're not calling your swap function but instead the swap defined in the standard library.

https://en.cppreference.com/w/cpp/algorithm/swap

This is the reason why using 'using namespace std' is a bad idea because if you didn't you'd have to explicitly call std::swap and you would have gotten an error otherwise in this case.

Answer 3

Thanks.