Memory Allocation Problem

Question

This question was asked in the written round of a job interview:

 #include<alloc.h>
 #define MAXROW 3
 #define MAXCOL 4

 main()
  {
    int (*p)[MAXCOL];
     p = (int (*)[MAXCOL]) malloc(MAXROW*(sizeof(*p)));
  }

How many bytes are allocated in the process?

To be honest, I did not answer the question. I did not understand the assignment to p.

Can anybody explain me what would be the answer and how it can be deduced?

Answer 1

It's platform dependent.

int (*p)[MAXCOL]; declares an pointer to an array of integers MAXCOL elements wide (MAXCOL of course is 4 in this case). One element of this pointer is therefore 4*sizeof(int) on the target platform.

The malloc statement allocates a memory buffer MAXROW times the size of the type contained in P. Therefore, in total, MAXROW*MAXCOL integers are allocated. The actual number of bytes will depend on the target platform.

Also, there's probably additional memory used by the C runtime (as internal bookeeping in malloc, as well as the various process initialization bits which happen before main is called), which is also completely platform dependant.

Answer 2

p is a pointer to an array of MAXCOL elements of type int, so sizeof *p (parentheses were redundant) is the size of such an array, i.e. MAXCOL*sizeof(int).

The cast on the return value of malloc is unnecessary, ugly, and considered harmful. In this case it hides a serious bug: due to missing prototype, malloc is assumed implicitly to return int, which is incompatible with its correct return type (void *), thus resulting in undefined behavior.