memory 分配是如何由 malloc 和 calloc function 在 Z0D61F8370CAD14D412F70B8D 中完成的？

Question

malloc() allocates a single block of memory whereas calloc() allocates multiple blocks of memory, each block with the same size.

How do they differ from each other in internal implementation?

I tried the below code and tried to figure out the difference between the internal implementation of malloc and calloc (considering the above-quoted text).

#include<stdio.h>
#include<stdlib.h>
int main()
{
   int *p=(int*)malloc(2*sizeof(int));
   p[0]=3;
   p[1]=4;
   printf("%d\t%d",p[0],p[1]);
   free(p);
   int *q=(int*)calloc(2,sizeof(int));
   q[0]=1;
   q[1]=2;
   printf("\n%d\t%d",q[0],q[1]);
   free(q);
   return 0;
}

But I didn't find any difference(from the output) between these two functions. What change should I do to understand the internal implementation of both functions? Here my meaning to the internal implementation is mentioned in the above-quoted text.

I hope you got my question.

Thanks

Answer 1

There is no big difference with the internal implementation perspective of both, or on most implementations calloc does use malloc followed by memset to zero the memory allocated as shown below example snippet. Refer to calloc implementation.

void *calloc(size_t n, size_t elem_size)
{
    const size_t nbytes = n * elem_size;
    void *p = malloc(nbytes);
    if (p != NULL) {
       memset(p, 0, nbytes);
    }
    return p;
}