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Does anyone know whether the (*) is optional around the function
pointer name when it is part of an argument list, e.g.

int f(int fn(int))
{
return fn(5);
}

?

Thank you for all your contributions. The C99 reference seems to
answer my question conclusively.

Richard Heathfield <> writes:
> said:
>> Does anyone know whether the (*) is optional around the function
>> pointer name when it is part of an argument list, e.g.

You mean parameter list, not argument list. An argument list is the
list of expressions in a function call.

>> int f(int fn(int))
>> {
>> return fn(5);
>> }
>>
>> ?
>
> Believe it or not, it *is* optional. 3.2.2.1 of C89 reads in part:
> "A function designator is an expression that has function type.
> Except when it is the operand of the sizeof operator /25/ or the
> unary & operator, a function designator with type ``function
> returning type '' is converted to an expression that has type
> ``pointer to function returning type .''"

Yes, but that's not the section that says it's optional. The text you
quoted says that the "*" is optional in a function call, not in a
function declaration.

The relevant paragraph is C99 6.7.5.3p8:

A declaration of a parameter as "function returning _type_" shall
be adjusted to "pointer to function returning _type_", as in
6.3.2.1.

> And I don't mind admitting that my initial reaction was "no". But I
> was wrong.

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Richard Heathfield <> writes:

> Keith Thompson said:
>
>> Richard Heathfield <> writes:
> <snip>
>>>
>>> Believe it or not, it *is* optional. 3.2.2.1 of C89 reads in
>>> part: "A function designator is an expression that has function
>>> type. Except when it is the operand of the sizeof operator /25/
>>> or the unary & operator, a function designator with type
>>> ``function returning type '' is converted to an expression that
>>> has type ``pointer to function returning type .''"
>>
>> Yes, but that's not the section that says it's optional. The text
>> you quoted says that the "*" is optional in a function call, not
>> in a function declaration.
>>
>> The relevant paragraph is C99 6.7.5.3p8:
>>
>> A declaration of a parameter as "function returning _type_"
>> shall be adjusted to "pointer to function returning _type_",
>> as in 6.3.2.1.
>
> Oops, nice catch. Thanks, Keith. The corresponding C89 quote is in
> 3.7.1, with the same wording, modulo the capital A and the
> cross-ref.

If it helps reduce the surprise factor, this is somewhat analogous to
the treatment of parameters that look as if they have an array type.

I too, only came across this very recently due to a typo and a
preference for keeping *s out of typedefs. I'd written:

typedef int func(int);

int some_function(func f) /* forgot the * by accident! */
{...}

and only found out that I should have been surprised that it worked
some time later when fixing something else! On the other hand, I
would not have been surprised that

typedef int vector[3];

int other_function(vector v) {...}

makes 'other_function' of type 'int(int *)' so there is some
similarity. Of course, the similarity stops when you *do* add the *.
It has no effect on 'some_function' but changes the type of
'other_function' to be 'int(int (*)[3])'.

--
Ben.