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 <<         


        

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.

Answer 2

You're passing a char* to the foo constructor. This can be implicitly converted to a boolean (as can all pointers) or to a std::string. From the compiler's point of view, the first conversion is "closer" than the second because it favours standard conversions (i.e. pointer to bool) over user provided conversions (the std::string(char*) constructor).

Answer 3

You're confusing two issues. One is that "blah" can be implicitly converted to a string and the other is that const char* can be implicitly converted into a boolean. It's very logical to see the compiler go to the implicit conversion which minimizes the total amount of conversions necessary.