Size of array function return size of int instead of array

Question

out put is always 8! I am wondering where this number comes from and how can I get the correct size of the size of the array which is 6?

#include <stdlib.h>
#include <stdio.h>

void cal(int a[])
{
    printf("the size of array is= %d\n", (int)(sizeof(a)/(sizeof)(int)));
}

int main ()
{
    int lis[]={1,2,3,4,5,6,7,8,9};
    cal(&lis[3]);
    return 0;
}

An array is a pointer to a datatype. `sizeof a` will result in the size of the pointer to the array. `sizeof *a` will derefernce the pointer and return the size of the datatype. `sizeof *a * numElems` will return the number of bytes in the array. — recursion.ninja, Sep 08 '13 at 19:00
@ppeterka66 no it's not. OP: I doubt the output is 8, much less that this compiles. — Luchian Grigore, Sep 08 '13 at 19:00
http://stackoverflow.com/questions/492384/how-to-find-the-sizeofa-pointer-pointing-to-an-array — Suvarna Pattayil, Sep 08 '13 at 19:02
@LuchianGrigore: Why wouldn't it compile? He's passing an `int *` to a function that's expecting an `int *`. — John Bode, Sep 08 '13 at 19:20

score 0 · Accepted Answer · answered Sep 08 '13 at 19:02

0

You can not calculate the size of the int array when all you've got is an int pointer. You will always get pointer size.

You can possibly try this(Not sure if that is a good idea):-

#define arrSz(a) (sizeof(a)/sizeof(*a))

answered Sep 08 '13 at 19:02

Rahul Tripathi

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Your `#define` works only if `a` is actually an array. In the OP's code, `a` is a pointer. – Keith Thompson Sep 08 '13 at 19:22
I see because we can get just an int pointer we can not have the size. It helps alot. Thanks. – Amir Nabaei Sep 09 '13 at 23:06
You are welcome. Don't forget to accept this, as, an answer if this helped you – Rahul Tripathi Sep 10 '13 at 03:03

Size of array function return size of int instead of array

1 Answers1