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

问题

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 deﬁne 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: deﬁnition of fv

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

回答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.

回答2:

It's probably mostly historical reasons. typedef was a relatively late addition to C, and was tacked onto the existing language (and caused a few problems for the parsing phase of compilers).

Also, a function definition has to define the names of the parameters, if any. A function type includes the function's return type and parameter types, but not its parameter names. For example, these:

void (int)
void (int x)
void (int y)

are three ways of writing the same function type. If you had:

typedef void func_t(int);

then this hypothetical definition:

func_t some_func { }

wouldn't define a name for its int parameter. I'm not sure how that could have been resolved in a reasonable manner. It would be possible, I suppose, but it was never done.

But the bottom line is probably just that Dennis Ritchie either didn't think it was worth the effort to define how a typedef could be used in a function definition, or he simply didn't think of it.

回答3:

Let me put a few words. Consider a statement:

typedef void F(int p1, char* p2);

This statement assigns name F to a function signature void (int, char*); This is definition of an alias to the function signature. After that the statement:

F fv;

tells that there is a function fv. It has the signature that was mentioned above and it has its body somewhere. Look at the C/C++ syntax of the function definition:

retType  funcName(params) { body }

There are actually 2 names used retType and funcName. None of them are the same to the name F from the initial typedef. The name F has meaning of both names. If language would allow something like:

F { body }

this will associate body with the function type. But this leads a problem:

The meaning of F would be not clear. Is it an "alias to the function signature" or is it a "name of the entry point into a code"?

Plus the syntax of the last example would be weird to millions of C/C++ programmers.

回答4:

The rule is as you quoted - typedef of function type shall not be used to define a function. In the 3rd line of the example, you are trying to define a function with function type F. This is not allowed by the standard.

EDIT
As you pointed out, I try to explain more on it.

For the 3rd line, if it were legal, then you could replace F with the typedef definition:
void fv { }(). This is not a legal definition or declaration in C++.

I think the key point is that typedef is just creating an alias for simplification, and you can replace your typedef type like replacement of #define during compilation.

来源：https://stackoverflow.com/questions/17848983/why-cant-a-typedef-of-a-function-be-used-to-define-a-function

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