Question

Consider the following excerpt from the safe bool idiom:

typedef void (Testable::*bool_type)() const;
operator bool_type() const;

Is it possible to declare the conversion function without the typedef? The following does not compile:

operator (void (Testable::*)() const)() const;

Answer 1

Ah, I just remembered the identity meta-function. It is possible to write

operator typename identity<void (Testable::*)() const>::type() const;

with the following definition of identity:

template <typename T>
struct identity
{
    typedef T type;
};

You could argue that identity still uses a typedef, but this solution is “good” enough for me.