const volatile char string not printing properly

Question

I have a very simple program which just outputs indefinitely a const pointer pointing to a const volatile char; it goes like this:

const volatile char* const str = "ABCDEFGHIJKL";

while(true) {
    cout << '\r' << str;
}

The problem is that when running this program, the output is 1. There is a way to get around this, which is outputting const_cast<char*>(str) instead of str.

But if I do const_cast<volatile char*>(str) the output is 1, just like before the cast, so I'm guessing the 1 output is caused by the volatile keyword, which is strange because I thought that volatile only makes the compiler avoid optimizations in that variable, which shouldn't change the value of it.

My question, therefore, is how in the world did that 1 came as the output.

NOTE:

I've tried compiling it with GCC in Ubuntu 16.04, and with MinGW in Windows 7, so the compiler is not the problem(I guess).

Answer 1

You got nuked by Implicit conversion Sequences (ICS). The C++ std::ostream facilities have no overloads for volatile types. ICS kicks in, and selects the overload for bool type (because pointer types, irrespective of cv qualifications are implicitly convertible to bool).

Hence you see 1... Change your output to std::boolalpha and you should see true instead.

Example:

#include <iostream>
#include <iomanip>

int main(){
    const volatile char* const str = "ABCDEFGHIJKL";
    std::cout << '\r' << str;
    std::cout << '\r' << std::boolalpha << str;
}

Prints:

1
true

Demo