«

"Flash Gordon" <> wrote in message
news:...
> Mike Wahler wrote, On 12/08/07 18:42:
>> "sumedh....." <> wrote in message
>> news: oups.com...
>>> On Aug 12, 9:21 pm, anon.a...@gmail.com wrote:
>>> sizeof(main) -> 1
>>> sizeof(main()) -> sizeof(int)?????
>>
>> The type of 'main' is a pointer type.
>
> Wrong. The time of main is a function,

Yes, its type is 'function'. But in a program,
the name (without parens) will yield an address.

-Mike

Mike Wahler wrote, On 12/08/07 20:34:
> "Flash Gordon" <> wrote in message
> news:...
>> Mike Wahler wrote, On 12/08/07 18:42:
>>> "sumedh....." <> wrote in message
>>> news: oups.com...
>>>> On Aug 12, 9:21 pm, anon.a...@gmail.com wrote:
>>>> sizeof(main) -> 1
>>>> sizeof(main()) -> sizeof(int)?????
>>> The type of 'main' is a pointer type.
>> Wrong. The time of main is a function,
>
> Yes, its type is 'function'. But in a program,
> the name (without parens) will yield an address.

N1124 section 6.3.2.1 para 4 it says
| A function designator is an expression that has function type.
| Except when it is the operand of the sizeof operator54) 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’’.

Further, the constraints section of the definition of the sizeof
operator says:
| The sizeof operator shall not be applied to an expression that
| has function type or ...

So as I said the type is function, not pointer to funtion, and a
diagnostic (warning or error) is *required*.
--
Flash Gordon

Mike Wahler wrote:
>
> "Flash Gordon" <> wrote in message

> > Because main is not a pointer and Mike was wrong.
>
> Huh? Are you saying that the name of a function (without parentheses)
> does not yeild a pointer value?

The only two times when an expression of function type is not
automatically converted to a pointer, are:
1 as an operand of &
2 as an operand of sizeof

Since sizeof is not defined for expressions of function type,
the only thing that you can do with an expression of function type
in a correct C program, is to derive a pointer from it.

--
pete

"Mike Wahler" <> writes:
[...]
> 'main' (without parentheses) yields the address of
> the function 'main()'.

In most contexts, yes.

> So sizeof(main) will give
> the size of a pointer to a function.

No, the operand of a sizeof operator is one of the contexts in which a
function name (more generally an expression of function type) is not
converted to a pointer. 'sizeof(main)' (or 'sizeof main'; the
parentheses are not necessary) is a constraint violation. <OT>gcc's
"-pedantic" option causes it to emit the required diagnostic.</OT>

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Aug 13, 2:35 am, pete <pfil...@mindspring.com> wrote:
> Mike Wahler wrote:
>
> > "Flash Gordon" <s...@flash-gordon.me.uk> wrote in message
> > > Because main is not a pointer and Mike was wrong.
>
> > Huh? Are you saying that the name of a function (without parentheses)
> > does not yeild a pointer value?
>
> The only two times when an expression of function type is not
> automatically converted to a pointer, are:
> 1 as an operand of &
> 2 as an operand of sizeof
>
> Since sizeof is not defined for expressions of function type,
> the only thing that you can do with an expression of function type
> in a correct C program, is to derive a pointer from it.
>
> --
> pete

pete would you like to explain by giving an example?
if i derive a pointer from it what would be the type of that pointer?

sumedh..... wrote:
>
> On Aug 13, 2:35 am, pete <pfil...@mindspring.com> wrote:
> > Mike Wahler wrote:
> >
> > > "Flash Gordon" <s...@flash-gordon.me.uk> wrote in message
> > > > Because main is not a pointer and Mike was wrong.
> >
> > > Huh? Are you saying that the name of a function (without parentheses)
> > > does not yeild a pointer value?
> >
> > The only two times when an expression of function type is not
> > automatically converted to a pointer, are:
> > 1 as an operand of &
> > 2 as an operand of sizeof
> >
> > Since sizeof is not defined for expressions of function type,
> > the only thing that you can do with an expression of function type
> > in a correct C program, is to derive a pointer from it.
> >
> > --
> > pete
>
> pete would you like to explain by giving an example?
> if i derive a pointer from it what would be the type of that pointer?

The type of (&main) is a pointer to a function
returning type int and taking no arguments.

--
pete

pete wrote:
>
> sumedh..... wrote:

> > > Since sizeof is not defined for expressions of function type,
> > > the only thing that you can do with an expression of function type
> > > in a correct C program, is to derive a pointer from it.
> > >
> > > --
> > > pete
> >
> > pete would you like to explain by giving an example?
> > if i derive a pointer from it what would
> > be the type of that pointer?
>
> The type of (&main) is a pointer to a function
> returning type int and taking no arguments.

Well actually, how many arguments depends on how main is defined.

--
pete

sumedh..... wrote:
> On Aug 13, 2:35 am, pete <pfil...@mindspring.com> wrote:
>> Mike Wahler wrote:
>>
>>> "Flash Gordon" <s...@flash-gordon.me.uk> wrote in message
>>>> Because main is not a pointer and Mike was wrong.
>>> Huh? Are you saying that the name of a function (without parentheses)
>>> does not yeild a pointer value?
>> The only two times when an expression of function type is not
>> automatically converted to a pointer, are:
>> 1 as an operand of &
>> 2 as an operand of sizeof
>>
>> Since sizeof is not defined for expressions of function type,
>> the only thing that you can do with an expression of function type
>> in a correct C program, is to derive a pointer from it.
>>
> pete would you like to explain by giving an example?
> if i derive a pointer from it what would be the type of that pointer?
>
Does this make it clearer?

typedef int (*Fn)(void);

int main( void ) {
Fn fn = main;
}

--
Ian Collins.

> Does this make it clearer?
>
> typedef int (*Fn)(void);
>
> int main( void ) {
> Fn fn = main;
>
> }
>
> --
> Ian Collins.

typedef int (*Fn)(void);
int main( void ) {
Fn fn = main;
printf("%d\n",sizeof(fn)); // o/ps 4 (that is size of pointer to
function returning int.)
printf("%d\n",sizeof(&fn)); // o/ps 4 (that is size of pointer to
function returning int.)
printf("%d\n",sizeof(&main)); // o/ps 4 (that is size of pointer to
function returning int.)
printf("%d\n",sizeof(main)); //o/ps 1 inspite of main & (&main) are
both the same thing.. pointer //to function returning int
printf("%d %d \n",fn,&fn); //o/ps addresses A & B
printf("%d %d \n",main,&main); //o/ps addresses A & A
system("PAUSE");
return 0;
}

sumedh..... wrote:
>> Does this make it clearer?
>>
>> typedef int (*Fn)(void);
>>
>> int main( void ) {
>> Fn fn = main;
>>
>> }
>>

*Please don't quote signatures.*

>
> typedef int (*Fn)(void);
> int main( void ) {
> Fn fn = main;

> printf("%d\n",sizeof(main)); //o/ps 1 inspite of main & (&main) are
> both the same thing.. pointer //to function returning int

Constraint violation, see 6.5.3.4.1

"The sizeof operator shall not be applied to an expression that has
function type..."

Did this even compile?

--
Ian Collins.