Why does the compiler force this conversion below into a bool value? [duplicate]

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