sizeof( function )

sizeof( function )

[Mihai Donțu]

Hi,

Not that is of great importance, but does anyone know why

int main( void )
{
	return printf( "%d\n", ( int )sizeof( main ) );
}

prints 1 ? :)

(sizeof( &main ) prints 4/8 as it should)

-- 
Mihai Donțu

Re: sizeof( function )

[Brian Dessent]

Mihai Don?u wrote:

> Not that is of great importance, but does anyone know why
> 
> int main( void )
> {
>         return printf( "%d\n", ( int )sizeof( main ) );
> }
> 
> prints 1 ? :)
> 
> (sizeof( &main ) prints 4/8 as it should)

In this context 'main' is a function type, i.e. a type that describes a
function of a given return type and a given number and type of arguments
(C99 §6.2.5 para. 20).  It is invalid to use sizeof() with a function
type expression (C99 §6.5.3.4 para. 1) and hence this code invokes
undefined behavior.  gcc gives a warning if you compile with -pedantic,
but otherwise it apparently just returns 1; it would be equally within
its rights to return 42 or to generate code to format the users hard
drive -- that is the nature of undefined behavior.

Brian

Re: sizeof( function )

[Segher Boessenkool]

>> Not that is of great importance, but does anyone know why
>>
>> int main( void )
>> {
>>         return printf( "%d\n", ( int )sizeof( main ) );
>> }
>>
>> prints 1 ? :)

> It is invalid to use sizeof() with a function
> type expression (C99 §6.5.3.4 para. 1) and hence this code invokes
> undefined behavior.  gcc gives a warning if you compile with -pedantic,
> but otherwise it apparently just returns 1;

See the description of -Wpointer-arith in the Fine Manual.


Segher

Re: sizeof( function )

[Mihai Donțu]

On Tuesday 12 June 2007 19:28, Brian Dessent wrote:
> Mihai Don?u wrote:
> 
> > Not that is of great importance, but does anyone know why
> > 
> > int main( void )
> > {
> >         return printf( "%d\n", ( int )sizeof( main ) );
> > }
> > 
> > prints 1 ? :)
> > 
> > (sizeof( &main ) prints 4/8 as it should)
> 
> In this context 'main' is a function type, i.e. a type that describes a
> function of a given return type and a given number and type of arguments
> (C99 §6.2.5 para. 20).  It is invalid to use sizeof() with a function
> type expression (C99 §6.5.3.4 para. 1) and hence this code invokes
> undefined behavior.  gcc gives a warning if you compile with -pedantic,
> but otherwise it apparently just returns 1; it would be equally within
> its rights to return 42 or to generate code to format the users hard
> drive -- that is the nature of undefined behavior.

  Unlike me, you really do know your C99 book :)

  Thanks a lot!

-- 
Mihai Donțu

Re: sizeof( function )

[Ian Lance Taylor]

Brian Dessent <brian@dessent.net> writes:

> Mihai Don?u wrote:
> 
> > Not that is of great importance, but does anyone know why
> > 
> > int main( void )
> > {
> >         return printf( "%d\n", ( int )sizeof( main ) );
> > }
> > 
> > prints 1 ? :)
> > 
> > (sizeof( &main ) prints 4/8 as it should)
> 
> In this context 'main' is a function type, i.e. a type that describes a
> function of a given return type and a given number and type of arguments
> (C99 §6.2.5 para. 20).  It is invalid to use sizeof() with a function
> type expression (C99 §6.5.3.4 para. 1) and hence this code invokes
> undefined behavior.  gcc gives a warning if you compile with -pedantic,
> but otherwise it apparently just returns 1; it would be equally within
> its rights to return 42 or to generate code to format the users hard
> drive -- that is the nature of undefined behavior.

Believe it or not, this is a documented gcc extension.

http://gcc.gnu.org/onlinedocs/gcc-4.2.0/gcc/Pointer-Arith.html

Ian

Re: sizeof( function )

[Martin Krischik]

Am 12.06.2007, 19:28 Uhr, schrieb Brian Dessent <brian@dessent.net>:

> Mihai Don?u wrote:
>
>> Not that is of great importance, but does anyone know why
>>
>> int main( void )
>> {
>>         return printf( "%d\n", ( int )sizeof( main ) );
>> }
>>
>> prints 1 ? :)
>>
>> (sizeof( &main ) prints 4/8 as it should)
>
> In this context 'main' is a function type, i.e. a type that describes a
> function of a given return type and a given number and type of arguments
> (C99 §6.2.5 para. 20).  It is invalid to use sizeof() with a function
> type expression (C99 §6.5.3.4 para. 1) and hence this code invokes
> undefined behavior.  gcc gives a warning if you compile with -pedantic,
> but otherwise it apparently just returns 1; it would be equally within
> its rights to return 42 or to generate code to format the users hard
> drive -- that is the nature of undefined behavior.

If it is invalid to use sizeof() with a function type expression - why not  
an make it an compile time error?

Martin

Re: sizeof( function )

[Nick Maclaren]

"Martin Krischik" <krischik@users.sourceforge.net> wrote:
>
> If it is invalid to use sizeof() with a function type expression - why not  
> an make it an compile time error?

That is what most compilers do.  However, gcc allows pointer arithmetic
on function pointers, to enable C code to poke through compiled code
(e.g. to disassemble code).  Given how few programmers ever do that
sort of thing, it could perfectly well issue a warning even without
-pedantic.  But it doesn't.

Note that, for C language reasons, it is virtually impossible to
separate the handling of sizeof(<type>) from allowing arithmetic on
values of type '<type> *'.


Regards,
Nick Maclaren,
University of Cambridge Computing Service,
New Museums Site, Pembroke Street, Cambridge CB2 3QH, England.
Email:  nmm1@cam.ac.uk
Tel.:  +44 1223 334761    Fax:  +44 1223 334679

Re: sizeof( function )

[Nick Maclaren]

I stand corrected.  When I last looked at the language accepted in
default (not -pedantic) mode, I didn't spot that one.  Sorry for
maligning gcc :-)

But, yes, I agree about using -pedantic.  Unless you have a real need
for gcc-specific extensions, compiling with that is always the Right
Thing To Do.  And -Wall and possibly -Wextra and ....


Regards,
Nick Maclaren,
University of Cambridge Computing Service,
New Museums Site, Pembroke Street, Cambridge CB2 3QH, England.
Email:  nmm1@cam.ac.uk
Tel.:  +44 1223 334761    Fax:  +44 1223 334679

Re: sizeof( function )

[Segher Boessenkool]

>> It is invalid to use sizeof() with a function
>> type expression (C99 '6.5.3.4 para. 1) and hence this code invokes
>> undefined behavior.  gcc gives a warning if you compile with 
>> -pedantic,
>> but otherwise it apparently just returns 1;

> But this is actually a BUG in gcc - look at 5.1.1.3.  The restriction
> that sizeof may not have an argument that is function type is a
> constraint, and therefore a breach requires a diagnostic.

There is no bug, as GCC prints a diagnostic just fine.
That's what -pedantic is for.


Segher

Re: sizeof( function )

[Brian Dessent]

Nick Maclaren wrote:

> But this is actually a BUG in gcc - look at 5.1.1.3.  The restriction
> that sizeof may not have an argument that is function type is a
> constraint, and therefore a breach requires a diagnostic.

It would only be a bug if there was no diagnostic in non-GNU mode, e.g.
-std=c89.  Otherwise the default is -std=gnu89 if not specified and as
the other poster pointed out, GNU extensions allow this and define the
return value as 1 for convenience.

Brian

Re: sizeof( function )

[Nick Maclaren]

> It is invalid to use sizeof() with a function
> type expression (C99 '6.5.3.4 para. 1) and hence this code invokes
> undefined behavior.  gcc gives a warning if you compile with -pedantic,
> but otherwise it apparently just returns 1;
It's not quite that simple - look at footnote 84 :-(

Both function designators ('function names') and sizeof are weird,
in that they look like a value and a function, but aren't quite.

But this is actually a BUG in gcc - look at 5.1.1.3.  The restriction
that sizeof may not have an argument that is function type is a
constraint, and therefore a breach requires a diagnostic.


Regards,
Nick Maclaren,
University of Cambridge Computing Service,
New Museums Site, Pembroke Street, Cambridge CB2 3QH, England.
Email:  nmm1@cam.ac.uk
Tel.:  +44 1223 334761    Fax:  +44 1223 334679