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why this code perfectly running and giving ouput 56

int main()
{
int *p=malloc(0);
*p=56;
printf("%d",*p);
return 0;
}

you can check malloc() manual in linux distros. if size in malloc() or calloc() call is zero then both of them may return NULL or some unique pointer value which may be further used by free().

return address and if u want to print malloc(o) than
main()
{
printf("%d",malloc(0));
}
o/p is some thing address

If size is zero, the return value depends on the particular library implementation (it may or may not be a null pointer), but the returned pointer shall not be dereferenced.

It will point to some relevant location and according to code :

int main()
{
int *p=malloc(0);
*p=56;
printf("%d",*p);
return 0;
}

The output will be 56 only.

If we remove the statement *p=56;, the some defined memory address is printed.

int main()
{
int *p=malloc(0);
*p=56;
printf("%d",*p);
return 0;
}

malloc return the void pointer and we know void pointer can be assigned in any kind of pointer without typecasting .
so here pointer p store the address of void pointer which returned by malloc, and at that address you are performing write operation by storing value 56,that's why it print 56.

Implementation dependent.
return a NULL pointer.
Same as ptr=NULL;
but realloc(ptr,new_size) works.