Why is my overloaded C++ constructor not called?

Question

I have a class like this one:

class Test{
public:
  Test(string value);
  Test(bool value);

};

If I create an object like this:

Test test("Just a test...");

The bool constructor is called!

Anyone knows why?

Thanks

Answer 1

The type of "Just a test..." is const char *, which can be implicitly converted to either bool or std::string. Because std::string isn't a built in type, the const char *s is converted to a bool. You can prevent that by explicitly converting the const char * to a std::string:

Test test(std::string("Just a test..."));

Answer 2

This is a well known C++ annoyance.

Your string literal has type of chat const[]. You've got two constructors, conversion sequences from char const[] to Test look like this:

1) char const[] -> char const* -> bool

2) char const[] -> char const* -> std::string

1) is a built-in standard conversion whereas 2) is a user-defined conversion. Built-in conversions have precedence over user defined conversions, thus your string literal gets converted more easily to bool than to std::string.