should be a simple question
i want to get the length of an int array
i tried sizeof(arr) <- returns 4
and i tried sizeof(arr)/sizeof(arr[0]) <- returns 1
the length is 9
when i do
printf("%d", sizeof(arr)/sizeof(arr[0])) it prints 9
but i want to do something like
size = sizeof(arr)/sizeof(arr[0]) and it gives me 1
what am i missing here ?
thanks all helpers
Question 6.21
Read all the links.
ok i read it but nothing to point me to the answer..
anyone?
sizeof pointer gives you sizeof the pointer. not sizeof what pointer is pointing to.
Thus
When array is passed to function, you're only passing address of first element.Code:int a[10]; // array int *p = a; // pointer p points to a sizeof(a) ; // gives sizeof(int) * 10 sizeof(p); // gives sizeof(int*)
ie. func( a ) is the same as func(&a[0]);
You should have posted what is 'arr'!
The best is to post minimal code that describes your problem.
Last edited by Bayint Naung; 05-15-2011 at 09:50 AM.
arr is 1,2,3,...,9
ok so by your example if i do
insertit should give me the size ?Code:sizeof(a)/sizeof(p)
it gives me 1 as the size
Post the damn code!.
Is arr declared as array or pointer?!
sizeof(a)/sizeof(p)? perhaps you want sizeof(a) / sizeof(a[0])
> printf("%d", sizeof(arr)/sizeof(arr[0])) it prints 9
This only works when the array is IN SCOPE.
Pointing to the array (even if it is in scope), or passing the array to another function (also making a pointer) means you end up applying sizeof to a pointer, and not to an array.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
At the risk of upstaging my friend Salem...Originally Posted by Salem
This happens because C does not actually know how big an array is, except in the scope where it's created. Outside that it's simply apointer to a block of memory that you, as a programmer, are required to treat correctly.
