Why is there an implicit type conversion from pointers to bool in C++?

Question

Consider the class foo with two constructors defined like this:

class foo
{
public:
    foo(const std::string& filename) {std::cout << "ctor 1" << std::endl;}
    foo(const bool some_flag = false) {std::cout << "ctor 2" << std::endl;}
};

Instantiate the class with a string literal, and guess which constructor is called?

foo a ("/path/to/file");

Output:

ctor 2

I don't know about you, but I don't find that the most intuitive behavior in programming history. I bet there is some clever reason for it, though, and I'd like to know what that might be?

Answer 1

It's very common in C to write this

void f(T* ptr) {
    if (ptr) {
        // ptr is not NULL
    }
}

You should make a const char* constructor.