C99中的哪个规定禁止通过typedef定义函数？

Question

I know that a function definition can't be done through typedef. For example:

typedef int f_t(int x, int y);

f_t fsum
{
    int sum;
    sum = x + y;
    return sum;
}

But I can't find the provision which forbids this definition in C99. Which provisions are related and how they forbid this definition?

Answer 1

This is immediately given in the clause describing function definitions:

6.9.1 Function definitions

Constraints

2 - 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⁾

141) The intent is that the type category in a function definition cannot be inherited from a typedef [...]

The way this works is that under declarators (6.7.5), the only function declarators (6.7.5.3) are those with parentheses after the identifier:

TD( parameter-type-list )
TD( identifier-list _opt )

So, if the function is defined via a typedef then it does not have the syntactic form of a function declarator and so is invalid by 6.9.1p2.

It may help to consider the syntactic breakdown of an attempted typedef function definition:

typedef int F(void);

F f {}
| | ^^-- compound-statement
| ^-- declarator
^-- declaration-specifiers

Note that the typedef type F is part of the declaration-specifiers and not part of the declarator , and so f (the declarator portion) does not have a function type.

Answer 2

Since you're asking about the intent of forbidding typedefs from being used in a function definition (in some comments), the C99 rationale has this to say about it:

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 ; } 

So it seems that the function defintion-by-typedef was disallowed because the definition didn't look enough like a function in that case.