Re: Basic C question

[Rene Herman <rene.herman@xxxxxxxxxxxx>]

On 03-08-08 17:38, rahul p wrote:

> On Sun, Aug 3, 2008 at 7:36 PM, Rene Herman
> <rene.herman@xxxxxxxxxxxx <mailto:rene.herman@xxxxxxxxxxxx>> wrote:

> > In C, except as an operand of sizeof or unary &, an expression
> > having function type is automatically converted from function type
> > to pointer to function.
> >
> > Or put differently: except for sizeof and unary &, 
> > "function_designator" and "&function_designator" are
> > interchangeable as expressions.
>
> Thanks Rene. Plz discard my earlier msg as I didn't get question then. 
> Yes it's the case with the fuctions.   Just a small detail. They aren't 
> interchangeable.  "&function_designator"  cannot be used in place of 
> "function_designator". Correct me if I am wrong.

As per my own followup, the specific thing I said was indeed wrong at 
that point but otherwise, with:

	void foo(void);

the expression "foo", other than for sizeof(foo) and &foo, has the same 
value as the expression &foo. Specifically, with:

	int (*bar)(void);

"bar = foo;" and "bar = &foo;" definitely make bar end up with the same 
value. Also, if we have a function that takes this function as a parameter:

	void func(void (*bar)(void));

calling func(foo) and func(&foo) do the same thing and neither should warn.

Moreover, with:

	typedef void (foo_t)(void);

We say that foo_t is a function returning void, and:

	void func(foo_t bar);

says that func is a function taking a function returning void as a 
parameter. When we actually call func(foo) though, foo is automatically 
converted into _a pointer_ to a function returning void, so func might 
as well have been declared as void func(foo_t *bar);

I expect that someone who enjoys reading the C standard more than I do 
could've given this answer in a single sentence (be it an unparseable 
one) but I believe what I should've said instead was just that

	void func(foo_t bar);

and

	void func(foo_t *bar);

are interchangeable.

Rene.

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