From: horst DOT kraemer AT t-online DOT de (Horst Kraemer)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: sizeof() but nothing to do with sizeof( some_structure )
Date: Wed, 01 Dec 1999 09:10:39 GMT
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On Wed, 01 Dec 1999 15:27:16 +1100, Michael Abbott aka frEk
<20014670 AT snetch DOT cpg DOT com DOT au> wrote:

On Wed, 01 Dec 1999 15:27:16 +1100, in comp.os.msdos.djgpp you wrote:

> Heya
> 
> Yep, sizeof() is a built in operator...
> 
> AFAIK, it's converted straight into a typeless token, ie if it's printf ("%i",
> sizeof(int)); it will be parsed to printf("%i", 4); and then that number will
> be converted to the appropriate type (in this case integer)...


No. It's vice versa. The compiler doesn't care if you wrote "%d" or
"%f" or whatever in your printf format string. It will blindly feed a
value of type size_t and not a "typeless token" (sizeof - although
usually evaluated at compile time - is _not_ a preprocessor
instruction) and it is _your_ task to specify the correct type.

Whenever you are sure that the value fits into an unsigned int then

	printf("%u\n", (unsigned)sizeof(mytype) );

is correct and portable to any ANSI compiler. Of course

	printf("%d\n", sizeof(mytype) );

usually happens to work, too, with DJGPP.


Regards
Horst