sizeof() confusion

[Dan SPALDING]

In article <9156 at latcs1.oz.au> jacob at latcs1.oz.au (Jacob L. Cybulski) writes:
-Being relatively new to C, I found a bit of a problem with a THINK C program
-that goes more or less like this :-
-
-typedef unsigned char Pattern[8];
-
-void foo (Pattern x)
-{
-/* 0 */	printf("%d\n", sizeof(Pattern);			/* prints 8 */
-/* 1 */	printf("%d\n", sizeof(x));			/* prints 4 */
-/* 2 */	printf("%d\n", sizeof(*x));			/* prints 1 */
-/* 3 */	printf("%d\n", sizeof((Pattern) x);		/* illegal */
-/* 4 */	printf("%d\n", sizeof(*((Pattern *) x));	/* prints 8 */
-}
-
In C, the base of an array is synonomous with a pointer to an element (the
base element) of that array.  Therefore, Pattern is 8 bytes, x is 4 (the size
of all pointers on the Mac) and *x will be 1 byte (an unsigned char).


Hope this helps.




-- 
------------------------------------------------------------------------
dan spalding -- microsoft systems/languages -- microsoft!dans
"there has not been a word invented that would convey my indifference to
that remark." - paraphrase from hawkeye pierce