My questions were:
> What is the output of:
>
> #include <stdio.h>
> int main(void) { printf("%d\n", sizeof(sizeof(int))); return 0; }
>
> Is this portable at all?
There was only *one* reply yet (via personal email), and, alas, it
was not correct... >:-)
Quoting Harbison/Steele's "C: A Reference Manual":
* 7.5.2 sizeof Operator
[...] In ANSI C the result of sizeof has the unsigned integer
type size_t defined in the header file stddef.h. Traditional C
implementations often use int or long as the result type [...].
* 11.2 NULL, ptrdiff_t, size_t, offsetof, wchar_t
[...] The type size_t is the unsigned integral type of the result
of the sizeof operator --- probably unsigned long; pre-ANSI
implementations use the (signed) type int for this purpose. [...]
Especially note the "fun" words like "often" or "probably". The
program prints 4 on Intel and HP-PA. It prints 8 on Alpha, but this
only by pure luck: %d expects an int (= 4 bytes) and was passed a
size_t = unsigned long = 8 bytes. If Alpha had another byte order,
it would have printed 0.
To make things even more confusing, consider integral promotion:
int i, j;
i = j * sizeof(...);
You could lose some of your bits during assignment here. I guess we
will all have much fun when Intel/HP release their Merced processor...
Cheers,
Sven
--
Sven Panne Tel.: +49/89/2178-2235
LMU, Institut fuer Informatik FAX : +49/89/2178-2211
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