Which provision in C99 forbids definition of a function through typedef?

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?

Related: [Why can't a typedef of a function be used to define a function?](http://stackoverflow.com/q/17848983/1009479) — Yu Hao, Oct 23 '13 at 08:04

ecatmur · Accepted Answer · 2013-10-21T17:54:11.837

4

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:

T D( parameter-type-list )
T D( 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.

edited Oct 21 '13 at 17:54

answered Oct 21 '13 at 17:16

ecatmur

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I have read this provision, but I can't understand how we get the intent. – junwanghe Oct 21 '13 at 17:20
Does that provision require a function declarator? It just requires a declarator. The 'fsum' is a legal declarator. – junwanghe Oct 21 '13 at 17:31
@junwanghe 'fsum' is a declarator, but without the declaration-specifier `f_t` it does not have a function type. – ecatmur Oct 21 '13 at 17:55
Thanks, this answer is helpful. – junwanghe Oct 22 '13 at 02:18

Michael Burr · Answer 2 · 2013-10-21T18:20:39.233

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.

Which provision in C99 forbids definition of a function through typedef?

2 Answers2

6.9.1 Function definitions