Thread: malloc function

Im learning the dynamic storage allocation functions and I have a question about why this is happening. I made a function my_malloc that takes my request in bytes as argument and should return a pointer to the block of memory. And I wanted to allocate space for an array of 10 numbers, so it should be 40 bytes right? After the call of my_malloc, when I check the sizeof of the new array it tells me that has 8 bytes, so having a length of 2 elements. When I initialize the array to 100 elements(more than what I asked for) there is no error or something..

Code:

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

void *my_malloc(int n);

int main(void)
{
    int *p, len;
    
    printf("my malloc request: %d bytes\n",(int)(10 * sizeof(int)));
    
    p = my_malloc(10 * sizeof(int));
    
    printf("Values after my_malloc call:\n\n"
           "size p: %d bytes\n", (int)sizeof(p));
    
    for(int i = 0; i < 100; i++) //Initializing all elements of p in 1
        p[i] = 1;
    
    len = sizeof(p)/sizeof(int);
    printf("len: %d\n", len);
    
    for(int i = 0; i < 100; i++) //printing the array
        printf("%d ", p[i]);
    printf("\n");
    
    len = sizeof(p)/sizeof(int);
    
    printf("len: %d\n", len);
    printf("size p: %d bytes\n", (int)sizeof(p));
    printf("\n");
    
    return 0;
}

void *my_malloc(int n)
{
    void *p = malloc(n);
    printf("Values inside my_malloc:\n\n"
            "n : %d\n", n);
    printf("size p : %d bytes\n\n",(int)sizeof(p));
    if (p == NULL) {
        printf("malloc failed in my_malloc function.\n");
        exit(EXIT_FAILURE);
    } else
        return p;
}

Can someone please explain me what's happening here? and/or what i'm doing wrong?