dynamically allocated variable size

Question

When I run this program, output is 4 bytes. ( I use a 64 bit compiler)

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

int main(void) {
    int *b;
    b=(int*)malloc(10*sizeof(int));
    printf("%d",sizeof(b));
    return 0;
}

But shouldn't the output be 40 bytes as I am dynamically allocating 40 bytes of space for 'b'?

A pointer is not an array. I got that. But is there any way that I can know how much bytes is allocated to b? — Surya Teja Vemparala, May 25 '16 at 20:42
It prints a size of the pointer itself. (Which is 4 bytes, 32 bits. Are you sure about 64-bit compiler?) There is no (portable and standard) way to get a size of heap-allocated block once it got allocated. — HolyBlackCat, May 25 '16 at 20:43
Btw, don't cast the result of `malloc()` in C, it's only needed in C++. — HolyBlackCat, May 25 '16 at 20:43
The pointer does not point to an array, but the the first element. Something like `int (*)[length]` is a pointer to an array. — too honest for this site, May 25 '16 at 20:56

score 0 · Answer 1 · edited May 25 '16 at 20:50

0

No; it should be the same as printf("%zu", sizeof(int *));, which on your implementation is 4, no matter what value you assign to b, since b has type int *, not int [10].

edited May 25 '16 at 20:50

HolyBlackCat

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answered May 25 '16 at 20:46

Jens

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dynamically allocated variable size

1 Answers1