c++ - Why can't a typedef of a function be used to define a function?

Question

From § 8.3.5.11 of ISO/IEC 14882:2011(E):

A typedef of function type may be used to declare a function but shall not be used to de?ne a function

The standard goes on to give this example:

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

What motivates this rule? It seems to limit the potential expressive usefulness of function typedefs.

Answer 1

Though this question is about C++, but since C++ inherits typedef and function pointer from C, so an explanation of the same question in C can be used in here. There's a formal explanation for C.

Rationale for International Standard - Programming Languages C §6.9.1 Function definitions

An argument list must be explicitly present in the declarator; it cannot be inherited from a typedef (see §6.7.5.3). That is to say, given the definition:

typedef int p(int q, int r);

the following fragment is invalid:

p funk // weird
{ return q + r ; }

Some current implementations rewrite the type of, for instance, a char parameter as if it were declared int, since the argument is known to be passed as an int in the absence of a prototype. The Standard requires, however, that the received argument be converted as if by assignment upon function entry. Type rewriting is thus no longer permissible.