Process finished with exit code 11 | Error during malloc

Question

Trying to understand memory allocation in C. Facing issue while trying to create two arrays using pointer to integers. Kindly have a look at the below code:

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

int main() {

    int *a;
    int *b;

    for (int i = 0; i<4;i++)
    {
        printf("Enter value \n");
        a[i]=(int *)malloc(sizeof(int));
        b[i]=(int *)malloc(sizeof(int));
        scanf("%d",&a[i]);
        scanf("%d",&b[i]);
    }
    for (int i =0;i<4;i++)
    {
      printf("%d = %x\n  ",a[i],&a[i]);

    }
    for (int i =0;i<4;i++)
    {
     printf("%d = %x\n  ",b[i],&b[i]);

    }

    return 0;
}

I am working with C11 on CLion. Facing below error on runtime. Can someone please explain what is wrong with this code ?

Enter value 

Process finished with exit code 11

"b is being shown NULL during debugging"

UPDATE: Tried on another IDE, where "a" itself is not being allocated any memory. It directly gives me segmentation fault.

UPDATE 2: Changing:

int *a;
int *b;

to

int *a = NULL;
int *b = NULL;

at least stops how this code is behaving. It gives me segmentation fault as soon as I try to allocate memory to a[i] (Which is wrong, now I get).

Answer 1

While the code do allocate memory for four int values, it does not work as you expect.

If we start from the beginning:

int *a;

This defines (and declares) a variable a which is a pointer to an int. You do not initialize it, which means its value is indeterminate (and will seem totally random). Dereferencing this (with e.g. a[i]) leads to undefined behavior.

Then

for (int i = 0; i<4;i++)
{
    a[i]=(int *)malloc(sizeof(int));
    ...
}

If you properly initialized a it would already point to memory for four int values. You could see a as an array, and you don't need to allocate each member of the array separately. This is indicated by the types, a[i] is of type int, not int *.

The "best" solution is to not use pointers or dynamic allocation at all, and instead use plain arrays:

int a[4];

for (int i = 0; i<4;i++)
{
    printf("Enter value \n");
    scanf("%d",&a[i]);  // No allocation needed, a[i] already exists
}

If you must use pointers and dynamic allocation, then allocate memory enough for four int elements, and make a point to the first byte:

int *a = malloc(sizeof(int) * 4);

for (int i = 0; i<4;i++)
{
    printf("Enter value \n");
    scanf("%d",&a[i]);  // No allocation needed, a[i] already exists
}

There are a few other problems as well, but start with these.

Answer 2

There is some confusion in your code:

• you do not allocate the arrays for a and b to point to. a and b are uninitialized, dereferencing them has undefined behavior. a shows as NULL in the debugger, but the C Standard does not guarantee that. a could have any value, including trap representations that would cause undefined behavior just by reading it.
• you store a pointer to int into a[i]: the compiler should issue a diagnostic about the type mismatch. It is unfortunate that this not be a fatal error IMHO, but ignoring compiler warnings is always a bad idea.
• the return value of malloc() is cast as (int *). This is necessary in C++, but considered bad style in C.
• the third argument in printf("%d = %x\n", a[i], &a[i]); is inconsistent with the conversion specifier: if you want to print the value in hex, use printf("%d = %x\n", a[i], a[i]);, if you want to print the address, use the %p conversion specifier: printf("%d at %p\n", a[i], (void*)&a[i]);

To play with malloc(), you should just allocate the int arrays, check for allocation success and free the memory after use:

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

int main() {
    int *a;
    int *b;

    a = malloc(sizeof(*a) * 4);
    b = malloc(sizeof(*b) * 4);
    if (a == NULL || b == NULL) {
        fprintf(stderr, "cannot allocate memory\n");
        exit(1);
    }

    for (int i = 0; i < 4; i++) {
        printf("Enter 2 values\n");
        if (scanf("%d%d", &a[i], &b[i]) != 2) {
            fprintf(stderr, "invalid input\n");
            exit(1);
        }
    }
    for (int i = 0; i < 4; i++) {
        printf("  %d = %x\n", a[i], a[i]);
    }
    for (int i = 0; i < 4; i++) {
        printf("  %d = %x\n", b[i], b[i]);
    }
    free(a);
    free(b);
    return 0;
}

Answer 3

You need to add these lines, because you need to create memory so that you can assign values later:

  int **a = (int **)malloc(sizeof(int *) * 4);
  int **b = (int **)malloc(sizeof(int *) * 4);

You are assigning pointer values to elements of a. Therefore a should be pointer to pointer. Complete code:

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

int main ()
{

  int **a = (int **)malloc(sizeof(int *) * 4);
  int **b = (int **)malloc(sizeof(int *) * 4);

  for (int i = 0; i < 4; i++)
    {
      printf ("Enter value \n");
      a[i] = (int *) malloc (sizeof (int));
      b[i] = (int *) malloc (sizeof (int));
      scanf ("%d", &a[i]);
      scanf ("%d", &b[i]);
    }
  for (int i = 0; i < 4; i++)
    {
      printf ("%d = %x\n  ", a[i], &a[i]);

    }
  for (int i = 0; i < 4; i++)
    {
      printf ("%d = %x\n  ", b[i], &b[i]);

    }

  return 0;
}

Output:

Enter value 
4
5
Enter value 
1
2
Enter value 
6
7
Enter value 
8
9
4 = 165d010
  1 = 165d018
  6 = 165d020
  8 = 165d028
  5 = 165d040
  2 = 165d048
  7 = 165d050
  9 = 165d058