GATE Overflow Test Series | Programming | Test 1 | Question: 1

Question

Consider the following C code fragment (… denote further statements):

int main()
{
    int my_array[30][40][50];
    init_array(&my_array);
    ...
}

Which of the following is a valid declaration for the function init_array? Mark all the correct options.

• A.  

  void init_array(int ** array[30][40][50]);

• B.  

  void init_array(int (* array)[30][40][50]);

• C.  

  void init_array(int *** array);

• D.  

  void init_array(int ** array[30][40]);

 

Comments

here ‘&my_array’ involves 4 levels of indirection, i.e. ****p=&my_array

Whereas just ‘my_array’ involves only 3 levels of indirection i.e. ***p=my_array

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@ reboot

int ****p = &my_array  is also not correct one. It is giving warning as incompatible type.

Answer 1

The argument being passed is a pointer to a 3-dimensional array of integers. So, only option B matches.

Comments

abhiTronix

void init_array_Y(int (* array)[30][40][50]){
  // incorrect implementation
  printf("%d\n",array[1][0][0]); //prints -1667136352 or some garbage value
}

arr is a pointer to int arr[30][40][50] type, instead of using print directly you are supposed to dereference first.
 

#include <stdio.h>

void fun(int *p){
    printf("%p\n",p); //prints address
    printf("%d\n",*p); //prints value 
}

int
main ()
{
  int i=10;
  fun(&i);
  printf("%p\n",&i);  //verify here, same address as passsed
  return 0;
}

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The reason I think for → “Why not C ?"

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int fun(int arr[]) → ID array

int fun(int * arr[]) → 2D array or Array of pointers

int fun(int ** arr[]) → 3D array or Array of pointers in which each element further points to an array of pointers.

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Now, In option C : they have given that : 

int fun(int *** array) → 4D array.

 

I think its clear now.

please correct me in case of any error → @Arjun, @gatecse

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int **

This has 3 dimensions but the first two dimensions are having pointers (addresses) while the innermost one is having integers.

Answer 2

Answer : Option B

As address of whole 3D array is being passed in calling function. In declaration we should have some pointer which accepts 3D array address. So, let’s go through all options : 

Option A : False, int ** array[30][40][50] this means it is a array of [30][40][50] integer double pointer. but here were are not accepting pointer, we are accepting address and for that single pointer should be there.

Option B : As previous stated we want to accept address through single pointer and in option B it is given so correct.

Option C : Same reasoning as option A, we want to accept address of whole 3D array not some pointer. Option c would be true if some double pointer is passed by calling function. 

Option D : Same reasoning as option A. 

Answer 3

See ,here what are we passing to the function , address of arr right!!! means the pointer pointing to the whole 3D array, So , i think option B is the correct answer

1. void init_array(int (* array)[30][40][50]); translates to this ->>>> array is a pointer to 3D array of int.