i know that unlike other variablesCode:printf("%p\n", array);
when we want to do some operation on an array
it uses the physical address of the array
but there are many cells in the array
does it prints every one of them??
does the array always defined as pointer ??
even if we make every cell as int
Last edited by transgalactic2; 10-14-2008 at 04:55 AM.
The compiler will pass an array as a pointer to the first element when passing it to a function. In this case, the function is printf, but it makes no difference to how the array is passed. It will show you the address of the first element of the array.
--
Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
> does it prints every one of them??
Try it for yourself... that'd be a no
> does the array always defined as pointer ?? even if we make every cell as int
Yes. But you should be casting to (void *) as "%p" requires (which may be implicit, but being explicit is always good).
Huh?
Because printf() allows a variable-number-of-arguments, I don't think the user's variable types can be checked against some expected prototype. Somewhere it is said that a pointer is a pointer is a pointer (its footprint, not "pointer arithmetic"), so stacking pointers to the printf() function will be safe, and always syntactically error free.
I don't think it's a size issue; it's just that %p prints "void *". It's not guaranteed to print anything else. Of course it does, for the reasons you mentioned.
