Implicit cast from const string to bool [duplicate]

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!

Answer 2

"1" is a string literal, that is array of char that gets converted to pointer and then to bool. Notice that this path is preferred to implicit construction of a temporary std::string object.

Answer 3

"1" is a string literal, which - when used as function argument - decays to a pointer of type const char*. As there is no overload for function foo taking a const char*, but there is a standard conversion from const char* to bool, it falls back to foo(bool). Note that a pointer value, when taken as a bool argument, is interpreted like somePtr==nullptr ? false : true.