Implicit cast from const string to bool [duplicate]

Question

I have the following code:

#include <iostream>
#include <string>

void foo(bool a)
{
        std::cout << "bool" << std::endl;
}

void foo(long long int a)
{
        std::cout << "long long int" << std::endl;
}

void foo(const std::string& a)
{
        std::cout << "string" << std::endl;
}

int main(int argc, char* args[])
{
        foo("1");
        return 0;
}

When executing I get this output:

bool

I would have expected as output:

string

Why does g++ 4.9 implicitly cast this string to bool?

Answer 1

Your compiler is interpreting the standard correctly. Yes, this is a tricky corner case that many interviewers ask so they appear smarter than they really are.

The route const char[2] (the formal type of the literal "1") to const char* to bool is a standard conversion sequence, since it uses exclusively built-in types.

Your compiler must favour that to a user-defined conversion sequence, viz. the std::string constructor from a const char*.

The presence of the overload void foo(long long int a) is a red herring.

You can rather elegantly work around this in C++11 by dropping your overload to bool, and writing

#include <type_traits>
template <
    typename Y,
    typename T = std::enable_if_t<std::is_same<Y, bool>{}>
>
void foo(Y)
{
  std::cout << "bool" << std::endl;
}

in its place. The compiler will then favour the std::string for const char[N] over the template (as that is one of the requirements of overload resolution). Nice!