sizeof a structure member
sig...@my-deja.com
structure (without using an additional typedef, et cetera)?
For example
struct blah {
int a;
char b[20];
}
I want the sizeof "blah.b", without having to declare a "blah",
nor having to create a separate type for "b". Is there a
way to do this? Intuitively this might be
sizeof(blah.b)
but that doesn't work.
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Before you buy.
Kaz Kylheku
>Is there a way to directly determine the sizeof a member of a
>structure (without using an additional typedef, et cetera)?
>
>For example
>
> struct blah {
> int a;
> char b[20];
> }
>
>I want the sizeof "blah.b", without having to declare a "blah",
>nor having to create a separate type for "b". Is there a
>way to do this?
sizeof ((struct blah *) 0)->b;
The expression under sizeof is not evaluated, so no null pointer dereference
actually takes place here.
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Morris Dovey
> Is there a way to directly determine the sizeof a member of a
> structure (without using an additional typedef, et cetera)?
>
> For example
>
> struct blah {
> int a;
> char b[20];
> }
>
> I want the sizeof "blah.b", without having to declare a "blah",
> nor having to create a separate type for "b". Is there a
> way to do this? Intuitively this might be
>
> sizeof(blah.b)
>
> but that doesn't work.
Try just using the constant 20 -- or you could #define BLAH_B_SIZE 20
and replace 20 where you need it with BLAH_B_SIZE -- or you could take
advantage of the typedef option (why not?).
Morris Dovey
West Des Moines, Iowa USA
Morris Dovey
> sizeof ((struct blah *) 0)->b;
>
> The expression under sizeof is not evaluated, so no null pointer dereference
> actually takes place here.
Kaz...
Most clever. (I guess that's what comes of reading the language specification.)
-hs-
>Kaz Kylheku wrote:
>
>> sizeof ((struct blah *) 0)->b;
>>
>> The expression under sizeof is not evaluated, so no null pointer
dereference
>> actually takes place here.
>
>Kaz...
>
>Most clever. (I guess that's what comes of reading the language
specification.)
Yes, it's fine.
Have a look to the ISO-C offsetof() macro in <stddef.h>
http://www.dinkum.com/htm_cl/stddef.html#offsetof
It is based on the same idea.
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Kaz Kylheku
>Morris Dovey a écrit dans le message <39F89162...@iedu.com>...
>>Kaz Kylheku wrote:
>>
>>> sizeof ((struct blah *) 0)->b;
>>>
>>> The expression under sizeof is not evaluated, so no null pointer
>dereference
>>> actually takes place here.
>>
>>Kaz...
>>
>>Most clever. (I guess that's what comes of reading the language
>specification.)
>
>
>Yes, it's fine.
>
>Have a look to the ISO-C offsetof() macro in <stddef.h>
>http://www.dinkum.com/htm_cl/stddef.html#offsetof
There is no ISO C macro in <stddef.h>. ISO C macros live in the ISO C document.
:)
The null pointer hack used by implementations of offsetof() is not
well-defined; it relies on pointer arithmetic involving the null pointer.
The null pointer hack in the trick I used is defined, because the pointer is
never evaluated.
Morris Dovey
> The null pointer hack used by implementations of offsetof() is not
> well-defined; it relies on pointer arithmetic involving the null pointer.
>
> The null pointer hack in the trick I used is defined, because the pointer is
> never evaluated.
Kaz...
I don't care if it /is/ a "hack". Don't even care that /you/ produced it. It's
beautiful! (I stole it.)
sig...@my-deja.com
k...@ashi.footprints.net wrote:
> >I want the sizeof "blah.b", without having to declare a "blah",
> >nor having to create a separate type for "b". Is there a
> >way to do this?
>
> sizeof ((struct blah *) 0)->b;
That is what I wanted, thanks.