typecheck code

typecheck code

[Sri Ram Vemulpali]

Hi all,

/*
 * Check at compile time that something is of a particular type.
 * Always evaluates to 1 so you may use it easily in comparisons.
 */
 #define typecheck(type,x) \
 ({      type __dummy; \
        typeof(x) __dummy2; \
        (void)(&__dummy == &__dummy2); \
        1; \
 })

#define typecheck_fn(type,function) \
({      typeof(type) __tmp = function; \
       (void)__tmp; \
})

Can anyone help me, explain the above code typecheck. How does
(void)(&__dummy == &__dummy2) evaluates to 1

I appreciate any explain.

-- 
Regards,
Sri.

typecheck code

[Sri Ram Vemulpali]

Hi all,

/*
 * Check at compile time that something is of a particular type.
 * Always evaluates to 1 so you may use it easily in comparisons.
 */
 #define typecheck(type,x) \
 ({      type __dummy; \
        typeof(x) __dummy2; \
        (void)(&__dummy == &__dummy2); \
        1; \
 })

#define typecheck_fn(type,function) \
({      typeof(type) __tmp = function; \
       (void)__tmp; \
})

Can anyone help me, explain the above code typecheck. How does
(void)(&__dummy == &__dummy2) evaluates to 1

I appreciate any explain.

-- 
Regards,
Sri.

Re: typecheck code

[Jamie Iles]

On Mon, Jan 31, 2011 at 12:03:37PM -0500, Sri Ram Vemulpali wrote:
> Hi all,
> 
> /*
>  * Check at compile time that something is of a particular type.
>  * Always evaluates to 1 so you may use it easily in comparisons.
>  */
>  #define typecheck(type,x) \
>  ({      type __dummy; \
>         typeof(x) __dummy2; \

So here we're creating __dummy of the macro specified type and __dummy2 
by using the GCC extensions to work out the type.

>         (void)(&__dummy == &__dummy2); \

This does a comparison and casts the result to void so we ignore the 
result at runtime (and the compiler can optimise it away), but at 
compile time, if the types don't match then we'll get:

	warning: comparison of distinct pointer types lacks a cast

>         1; \

and here we always return 1.

>  })
> 
> #define typecheck_fn(type,function) \
> ({      typeof(type) __tmp = function; \
>        (void)__tmp; \
> })
> 
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

It's not that comparison that is evaluating to 1, it's the line 
immediately after.  The GCC statement expression extensions [1] mean 
that the "1; \" at the end will always be the value that the macro is 
evaluated to.

Jamie

1. http://gcc.gnu.org/onlinedocs/gcc/Statement-Exprs.html

Re: typecheck code

[Geert Uytterhoeven]

On Mon, Jan 31, 2011 at 18:03, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
>  * Check at compile time that something is of a particular type.
>  * Always evaluates to 1 so you may use it easily in comparisons.
>  */
>  #define typecheck(type,x) \
>  ({      type __dummy; \
>        typeof(x) __dummy2; \
>        (void)(&__dummy == &__dummy2); \
>        1; \
>  })
>
> #define typecheck_fn(type,function) \
> ({      typeof(type) __tmp = function; \
>       (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

It does not rely on the equation evaluating to 1. That result is
expicitly unused,
witness the cast to void.
It does rely on gcc complaining if you compare two pointers that don't
point to the same type.

Gr{oetje,eeting}s,

                        Geert

--
Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@linux-m68k.org

In personal conversations with technical people, I call myself a hacker. But
when I'm talking to journalists I just say "programmer" or something like that.
                                -- Linus Torvalds

Re: typecheck code

[Rajat Sharma]

> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

Its not this comparison getting evaluated to 1, but last expression
"1; \" which is forcibly returning 1 in every case. Since it is just
compile time warning and should not effect the program execution, this
macro is always returning 1 at run time. To me it just valuable for
coding style where you do some action only if type matches, but in
real sense you take action anyways.

Rajat

On Mon, Jan 31, 2011 at 10:33 PM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
>  * Check at compile time that something is of a particular type.
>  * Always evaluates to 1 so you may use it easily in comparisons.
>  */
>  #define typecheck(type,x) \
>  ({      type __dummy; \
>        typeof(x) __dummy2; \
>        (void)(&__dummy == &__dummy2); \
>        1; \
>  })
>
> #define typecheck_fn(type,function) \
> ({      typeof(type) __tmp = function; \
>       (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1
>
> I appreciate any explain.
>
> --
> Regards,
> Sri.
>
> _______________________________________________
> Kernelnewbies mailing list
> Kernelnewbies@kernelnewbies.org
> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
>

typecheck code

[Rajat Sharma]

> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

Its not this comparison getting evaluated to 1, but last expression
"1; \" which is forcibly returning 1 in every case. Since it is just
compile time warning and should not effect the program execution, this
macro is always returning 1 at run time. To me it just valuable for
coding style where you do some action only if type matches, but in
real sense you take action anyways.

Rajat

On Mon, Jan 31, 2011 at 10:33 PM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
> ?* Check at compile time that something is of a particular type.
> ?* Always evaluates to 1 so you may use it easily in comparisons.
> ?*/
> ?#define typecheck(type,x) \
> ?({ ? ? ?type __dummy; \
> ? ? ? ?typeof(x) __dummy2; \
> ? ? ? ?(void)(&__dummy == &__dummy2); \
> ? ? ? ?1; \
> ?})
>
> #define typecheck_fn(type,function) \
> ({ ? ? ?typeof(type) __tmp = function; \
> ? ? ? (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1
>
> I appreciate any explain.
>
> --
> Regards,
> Sri.
>
> _______________________________________________
> Kernelnewbies mailing list
> Kernelnewbies at kernelnewbies.org
> http://lists.kernelnewbies.org/mailman/listinfo/kernelnewbies
>

Re: typecheck code

[Manish Katiyar]

On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
>  * Check at compile time that something is of a particular type.
>  * Always evaluates to 1 so you may use it easily in comparisons.
>  */
>  #define typecheck(type,x) \
>  ({      type __dummy; \
>        typeof(x) __dummy2; \
>        (void)(&__dummy == &__dummy2); \
>        1; \
>  })
>
> #define typecheck_fn(type,function) \
> ({      typeof(type) __tmp = function; \
>       (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

Infact I think it will never return 1, since the addresses of __dummy1
and __dummy2 have to be different (off by 4 or 8). As pointed out it
is the next line that always returns 1. The purpose of this line is to
throw away warnings like "Incompatible pointer comparison" or
something like that (haven't tried :-)) incase there is a mismatch.

-- 
Thanks -
Manish

typecheck code

[Manish Katiyar]

On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
> ?* Check at compile time that something is of a particular type.
> ?* Always evaluates to 1 so you may use it easily in comparisons.
> ?*/
> ?#define typecheck(type,x) \
> ?({ ? ? ?type __dummy; \
> ? ? ? ?typeof(x) __dummy2; \
> ? ? ? ?(void)(&__dummy == &__dummy2); \
> ? ? ? ?1; \
> ?})
>
> #define typecheck_fn(type,function) \
> ({ ? ? ?typeof(type) __tmp = function; \
> ? ? ? (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1

Infact I think it will never return 1, since the addresses of __dummy1
and __dummy2 have to be different (off by 4 or 8). As pointed out it
is the next line that always returns 1. The purpose of this line is to
throw away warnings like "Incompatible pointer comparison" or
something like that (haven't tried :-)) incase there is a mismatch.

-- 
Thanks -
Manish

Re: typecheck code

[Sri Ram Vemulpali]

Thanks for all explanation. It really helped to understand.

Sri

On Mon, Jan 31, 2011 at 1:03 PM, Manish Katiyar <mkatiyar@gmail.com> wrote:
> On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
> <sri.ram.gmu06@gmail.com> wrote:
>> Hi all,
>>
>> /*
>>  * Check at compile time that something is of a particular type.
>>  * Always evaluates to 1 so you may use it easily in comparisons.
>>  */
>>  #define typecheck(type,x) \
>>  ({      type __dummy; \
>>        typeof(x) __dummy2; \
>>        (void)(&__dummy == &__dummy2); \
>>        1; \
>>  })
>>
>> #define typecheck_fn(type,function) \
>> ({      typeof(type) __tmp = function; \
>>       (void)__tmp; \
>> })
>>
>> Can anyone help me, explain the above code typecheck. How does
>> (void)(&__dummy == &__dummy2) evaluates to 1
>
> Infact I think it will never return 1, since the addresses of __dummy1
> and __dummy2 have to be different (off by 4 or 8). As pointed out it
> is the next line that always returns 1. The purpose of this line is to
> throw away warnings like "Incompatible pointer comparison" or
> something like that (haven't tried :-)) incase there is a mismatch.
>
> --
> Thanks -
> Manish
>



-- 
Regards,
Sri.

typecheck code

[Sri Ram Vemulpali]

Thanks for all explanation. It really helped to understand.

Sri

On Mon, Jan 31, 2011 at 1:03 PM, Manish Katiyar <mkatiyar@gmail.com> wrote:
> On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
> <sri.ram.gmu06@gmail.com> wrote:
>> Hi all,
>>
>> /*
>> ?* Check at compile time that something is of a particular type.
>> ?* Always evaluates to 1 so you may use it easily in comparisons.
>> ?*/
>> ?#define typecheck(type,x) \
>> ?({ ? ? ?type __dummy; \
>> ? ? ? ?typeof(x) __dummy2; \
>> ? ? ? ?(void)(&__dummy == &__dummy2); \
>> ? ? ? ?1; \
>> ?})
>>
>> #define typecheck_fn(type,function) \
>> ({ ? ? ?typeof(type) __tmp = function; \
>> ? ? ? (void)__tmp; \
>> })
>>
>> Can anyone help me, explain the above code typecheck. How does
>> (void)(&__dummy == &__dummy2) evaluates to 1
>
> Infact I think it will never return 1, since the addresses of __dummy1
> and __dummy2 have to be different (off by 4 or 8). As pointed out it
> is the next line that always returns 1. The purpose of this line is to
> throw away warnings like "Incompatible pointer comparison" or
> something like that (haven't tried :-)) incase there is a mismatch.
>
> --
> Thanks -
> Manish
>



-- 
Regards,
Sri.

Re: typecheck code

[Dave Hylands]

Hi Sri,

On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
>  * Check at compile time that something is of a particular type.
>  * Always evaluates to 1 so you may use it easily in comparisons.
>  */
>  #define typecheck(type,x) \
>  ({      type __dummy; \
>        typeof(x) __dummy2; \
>        (void)(&__dummy == &__dummy2); \
>        1; \
>  })
>
> #define typecheck_fn(type,function) \
> ({      typeof(type) __tmp = function; \
>       (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1
>
> I appreciate any explain.

If dummy and dummy2 are of different types, then when you try and do a
pointer comparison (&dummy == &dummy2) it will produce a compiler
warning/error.

The actual comparison will always fail, but it doesn't matter since
the results aren't used.

typecheck always returns 1.

Dave Hylands

typecheck code

[Dave Hylands]

Hi Sri,

On Mon, Jan 31, 2011 at 9:03 AM, Sri Ram Vemulpali
<sri.ram.gmu06@gmail.com> wrote:
> Hi all,
>
> /*
> ?* Check at compile time that something is of a particular type.
> ?* Always evaluates to 1 so you may use it easily in comparisons.
> ?*/
> ?#define typecheck(type,x) \
> ?({ ? ? ?type __dummy; \
> ? ? ? ?typeof(x) __dummy2; \
> ? ? ? ?(void)(&__dummy == &__dummy2); \
> ? ? ? ?1; \
> ?})
>
> #define typecheck_fn(type,function) \
> ({ ? ? ?typeof(type) __tmp = function; \
> ? ? ? (void)__tmp; \
> })
>
> Can anyone help me, explain the above code typecheck. How does
> (void)(&__dummy == &__dummy2) evaluates to 1
>
> I appreciate any explain.

If dummy and dummy2 are of different types, then when you try and do a
pointer comparison (&dummy == &dummy2) it will produce a compiler
warning/error.

The actual comparison will always fail, but it doesn't matter since
the results aren't used.

typecheck always returns 1.

Dave Hylands

Re: typecheck code

[julie Sullivan]

>If dummy and dummy2 are of different types, then when you try and do a
>pointer comparison (&dummy == &dummy2) it will produce a compiler
>warning/error.

O--ooohhh...
I came across this ages ago when I was trying to unravel the jiffies
wraparound macros and I've been scratching my head ever since trying
to figure out what this line _really_  did. A bit arcane, isn't it?
Thanks a lot Dave and everyone for explaining.

Cheers
Julie

typecheck code

[julie Sullivan]

>If dummy and dummy2 are of different types, then when you try and do a
>pointer comparison (&dummy == &dummy2) it will produce a compiler
>warning/error.

O--ooohhh...
I came across this ages ago when I was trying to unravel the jiffies
wraparound macros and I've been scratching my head ever since trying
to figure out what this line _really_  did. A bit arcane, isn't it?
Thanks a lot Dave and everyone for explaining.

Cheers
Julie