Why does a quoted string match bool method signature before a std::string?

Question

Given the following methods:

// Method 1
void add(const std::string& header, bool replace);

//Method 2
void add(const std::string& name, const std::         


        

Answer 1

My guess is the conversion from pointer to bool is an implicit primitive type conversion, where the conversion to std::string requires the call of a constructor and the construction of a temporary.

Answer 2

Pointers have an implicit conversion to bool. Perhaps you have seen the following:

void myFunc(int* a)
{
    if (a)
        ++(*a);
}

Now, in C++, implicit conversions between built-in types take precedence over conversions between class-types. So for example, if you had a class:

class Int
{
public:
    Int(int i) {}
}

And you overloaded a function for long and Int:

void test(long n) {cout << "long";}
void test(Int n) {cout << "Int";}

You'll see that the following code calls the long overload:

int i;
test(i);

Answer 3

In your case you have has overloaded functions. Overloading resolution occurs according to Section 13.3.

C++03 13.3.3.2/2:

When comparing the basic forms of implicit conversion sequences (as defined in 13.3.3.1)
— a standard conversion sequence (13.3.3.1.1) is a better conversion sequence than a user-defined conversion sequence or an ellipsis conversion sequence, and
— a user-defined conversion sequence (13.3.3.1.2) is a better conversion sequence than an ellipsis conversion sequence (13.3.3.1.3).

Conversion pointer into bool is a standard conversion. Conversion pointer into std::string is a user-defined conversion.

4.12 Boolean conversions An rvalue of arithmetic, enumeration, pointer, or pointer to member type can be converted to an rvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true.