Are there cases where a typedef is absolutely necessary?

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.