Conversion operator + conversion constructor = unintuitive behavior?

Question

I don\'t understand why the code below prints struct Value instead of int (which implies the conversion constructor is converting to Value

Answer 1

In the first example Visual Studio is incorrect; the call is ambiguous. gcc in C++03 mode prints:

source.cpp:21:34: error: call of overloaded 'Value(Convertible)' is ambiguous
source.cpp:21:34: note: candidates are:
source.cpp:9:5: note: Value::Value(int)
source.cpp:6:8: note: Value::Value(const Value&)

Recall that a copy constructor is implicitly defaulted. The governing paragraph is 13.3.1.3 Initialization by constructor [over.match.ctor]:

When objects of class type are direct-initialized [...], overload resolution selects the constructor. For direct-initialization, the candidate functions are all the constructors of the class of the object being initialized.

In the second example, deleted functions participate equally in overload resolution; they only affect compilation once overloads have been resolved, when a program that selects a deleted function is ill-formed. The motivating example in the standard is of a class that can only be constructed from floating-point types:

struct onlydouble {
  onlydouble(std::intmax_t) = delete;
  onlydouble(double);
};

Answer 2

I've tried your code (on the Visual Studio version only).

Since you have a built-in copy-CTOR, I guess your main is equal to:

int main()
{
    try { Value w((struct Value)(Convertible())); }
    catch (char const *s) { cerr << s << endl; }
}

The compiler has chosen to use your copy CTOR, rather than Value(int).

Changing it to:

int main()
{
    try { Value w((int)(Convertible())); }
    catch (char const *s) { cerr << s << endl; }
}

It printed "int".