Question

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Why is address of char data not displayed?

I was experimenting with ampersand operator and got stuck at this program :

#include<iostream>
using namespace std;

int main() {
    char i='a';
    cout<<&i;
    return 1;
}

I was expecting the address of variable i as the output but instead the output came as the value of variable i itself.

Can anybody explain what just happened? Thanx in advance.

Answer 1

That’s because cout::operator<< has an overload for const char*. You’ll need an explicit cast to print the address:

cout<<static_cast<void*>(&i);

This will call the overload with void* as parameter, which is the one used to print addresses.

Also note that your code runs into undefined behavior. You only have a single char there, and the overload expects a null-terminated C-string.