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[QUOTE="Guest, post: 3872170"]
no. Think about it from the compiler's point of view. It actually
has enough information to calculate the value of "sizeof *foo"
without dereferencing foo. It doesn't need to know the value
of foo but only the type of foo. And since it's the compiler
it can simply look up the type of foo in an internal table
and substitute the size of type of foo. Which it also *must*
know since its the compiler (baring incomplete definitions
etc.).

calloc (sizeof *foo);

might well compile to

load R1 #sizeof(struct-tm)
  call calloc

where #sizeof(struct-tm) is some sort of internal compiler
directive that in turn expands to something like

load R1 LIT 24

in in a more compact assembler

ld r1 #24

In fact not only is the compiler *able* to do this
but it is *required* to do this. sizeof is not
a function call but an operator

printf ("%ul\n", sizeof (* ((int*)0) ));

won't crash (assuming I got all me brackets right!)

<snip>

--
Nick Keighley

But ye have set at nought all my counsel, and would none of my
reproof:
I also will laugh at your calamity; I will mock when your fear
cometh;
[Proverbs 1:25]
[/QUOTE]

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