Can I use a function typedef in function definitions?

Question

I've defined:

typedef int FunkyFunc(int x);

Now, I would like to be able to use this typedef in the definition of functions of type FunkyFunc, e.g.

FunkyFunc f {
    return 2*x;
}

or

FunkyFunc f(int x) {
    return 2*x;
}

or

FunkyFunc f(x) {
    return 2*x;
}

can I do something similar to any of the above? None of them seems to compile.

Answer 1

This is ill-formed in C++ from the draft standard section 8.3.5 Functions paragraph 10 says:

A typedef of function type may be used to declare a function but shall not be used to define a function (8.4). [ Example:

typedef void F();
F fv; // OK: equivalent to void fv();
F fv { } // ill-formed
void fv() { } // OK: definition of fv

—end example ][...]

We can see in C this is also specifically forbidden from the C99 draft standard section 6.9.1 Function definitions says:

The identifier declared in a function definition (which is the name of the function) shall have a function type, as specified by the declarator portion of the function definition.¹⁴¹⁾

and footnote 141 says:

The intent is that the type category in a function definition cannot be inherited from a typedef:

and has the following example:

typedef int F(void); // type F is ‘‘function with no parameters
                     // returning int’’
F f, g; // fand g both have type compatible with F
F f { /* ... */ } // WRONG: syntax/constraint error
[...]

C11 says the same things except the footnote is 162.

Answer 2

No. You can use the typedef to declare a function:

FunkyFunc f;

but a function definition must be written with a function-style declarator.

_{Note: This is certainly the case in C++, and I'm fairly sure C is the same in this regard; but it would be better if you chose a single language to ask about, since there can be significant differences between C and C++ even where you might imagine they'd be the same.}