Why is the address of this volatile variable always at 1?

Question

I wanted to inspect the address of my variable

volatile int clock; cout << &clock; 

But it always says that x is at address 1. Am i doing something wrong??

Answer 1

iostreams will cast most pointers to void * for display - but no conversion exists for volatile pointers. As such C++ falls back to the implicit cast to bool. Cast to void* explicitly if you want to print the address:

std::cout << (void*)&clock; 

Answer 2

There's an operator<< for const void*, but there's no operator<< for volatile void*, and the implicit conversion will not remove volatile (it won't remove const either).

As GMan says, the cv-qualification of the type pointed to should be irrelevant to the business of printing an address. Perhaps the overload defined in 27.7.3.6.2 should be operator<<(const volatile void* val);, I can't immediately see any disadvantage. But it isn't.

#include <iostream>  void foo(const void *a) {     std::cout << "pointer\n"; }  void foo(bool a) {     std::cout << "bool\n"; }  int main() {     volatile int x;     foo(&x);     std::cout << &x << "\n";     int y;     foo(&y);     std::cout << &y << "\n";     void foo(volatile void*);     foo(&x); }  void foo(volatile void *a) {     std::cout << "now it's a pointer\n"; } 

Output:

bool 1 pointer 0x22cd28 now it's a pointer