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

Answer 1

This is because there is no overload for operator << that takes a pointer to volatile, and there is no pointer conversion that could satisfy it.

According to C++ standard,

for any type T, pointer to T, pointer to const T, and pointer to volatile T are considered distinct parameter types, as are reference to T, reference to const T, and reference to volatile T.

Operator << has no overload for pointers to non-static member, pointers to volatile, or function pointers, so attempting to output such objects invokes implicit conversion to bool.