malloc_usable_size() returns the wrong size

Question

I want to know the size allocated by malloc.
I have written the source code below.

test.c

#include <stdio.h>
#include <stdint.h>
#include <malloc.h>
void main(void)
{
    uint8_t *test;

    test = (uint8_t *)malloc(sizeof(uint8_t)*4);
    printf("sizeof(test) = %d\n",malloc_usable_size(test));

    free(test); 
}  

I expected size to be 4.
But the result is 12.

sizeof(test) = 12

Can you tell me what's wrong?
I hope that size 4 correctly comes out.

Answer 1

malloc_usable_size(test)

The value returned by above function is not fixed as you requested. it may be greater than the requested size of the allocation depending upon the cpu byte ordering and alignment. this is totally depend upon the underlaying implementation.