Using & operator with char data type

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.

The above program gives the expected output for int, float, double and long data types. — Maverick Snyder, Dec 09 '12 at 15:24

score 4 · Accepted Answer · answered Dec 09 '12 at 15:23

4

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.

answered Dec 09 '12 at 15:23

Luchian Grigore

I disagree about that the behavior is undefined. The shift operator behaves exactly as it is documented, it's the data that is corrupted :) – StoryTeller - Unslander Monica Dec 09 '12 at 15:27
@DimaRudnik there's no shift operator. – Luchian Grigore Dec 09 '12 at 15:27
Isn't that what the operator is called, shift left? – StoryTeller - Unslander Monica Dec 09 '12 at 15:29
@DimaRudnik it's not shifting bits, is it? I call it the stream operator, but come to think of it, I'm not sure that's the name. – Luchian Grigore Dec 09 '12 at 15:32
@DimaRudnik you're right, it's called the left shift operator. My bad :). The behavior is undefined because `cout` will continue printing until it finds a `'\0'`, but you only own the first `char`. Reading the next ones is illegal. – Luchian Grigore Dec 09 '12 at 15:36
For streams it's typically called the stream insertion operator; it has zilch to do with bit shifting. The behavior is undefined the instance that `std::ostream::operator<<(const char *)` attempts to access the character after `a` in `&i` (and this attempted access is exactly what will happen). – David Hammen Dec 09 '12 at 15:55

1 Answers1