Difference between “pointer to int” and “pointer to array of ints”

Question

int main()
{
    int (*x)[5];                 //pointer to an array of integers
    int y[6] = {1,2,3,4,5,6};    //array of integers
    int *z;                              


        

Answer 1

#include<stdio.h>

int main(void)
{
    int (*x)[6];                 //pointer to an array of integers
    int y[6] = {11,22,33,44,55,66};    //array of integers
    int *z;                      //pointer to integer
    int i;

    z = y;
    for(i = 0;i<6;i++)
        printf("%d ",z[i]);
    printf("\n");

    x = &y;

    for(int j = 0;j<6;j++)
        printf("%d ",*(x[0]+j));

    return 0;
}

//OUTPUT::

11 22 33 44 55 66

11 22 33 44 55 66

Pointer to an array are best suitable for multi-dimensional array. but in above example we used single dimension array. so, in the second for loop we should use (x[0]+j) with * to print the value. Here, x[0] means 0th array. And when we try to print value using printf("%d ",x[i]); you will get 1st value is 11 and then some garbage value due to trying to access 1st row of array and so on.

Answer 2

One should understand the internal representation of (*x)[i]. Internally, it is represented as *((*x)+i), which is nothing but the ith element of the array to which x is pointing. This is also a way to have a pointer pointing to 2d array. The number of rows is irrelevant in a 2d array.

For example:

int arr[][2]={{1,2},{3,4}};
int (*x)(2);
x=arr; /* Now x is a pointer to the 2d array arr.*/

Here x is pointing to a 2d array having 2 integer values in all columns, and array elements are stored contiguously. So (*x)[0] will print arr[0][0] (which is 1), (*x)[1] will print the value of arr[0][1] (which is 2) and so on. (*x+1)[0] will print the value of arr[1][0] (3 in this case) (*x+1)[1] will print the value of arr[1][1] (4 in this case) and so on.

Now, a 1d array could be treated as nothing but a 2d array having only one row with as many columns.

int y[6] = {1,2,3,4,5,6};
int (*x)[6];
x =y;

This means x is a pointer to an array having 6 integers. So (*x)[i] which is equivalent to *((*x)+i) will print ith index value of y.

Answer 3

The short answer: There is a difference, but your example is flawed.

The long answer:

The difference is that int* points to an int type, but int (*x)[6] points to an array of 6 ints. Actually in your example,

x = y;

is undefined** behavior, you know these are of two different types, but in C you do what you want. I'll just use a pointer to an array of six ints.

Take this modified example:

int (*x)[6];                 //pointer to an array of integers
int y[6] = {1,2,3,4,5,6};    //array of integers
int *z;                      //pointer to integer
int i;

z = y;
for(i = 0;i<6;i++)
    printf("%d ",z[i]);

x = y; // should be x = &y but leave it for now!

for(i = 0;i<6;i++)
    printf("%d ",x[i]); // note: x[i] not (*x)[i]

First,

1 2 3 4 5 6

Would be printed. Then, we get to x[0]. x[0] is nothing but an array of 6 ints. An array in C is the address of the first element. So, the address of y would be printed, then the address of the next array in the next iteration. For example, on my machine:

1 2 3 4 5 6 109247792 109247816 109247840 109247864 109247888 109247912

As you can see, the difference between consecutive addresses is nothing but:

sizeof(int[6]) // 24 on my machine!

In summary, these are two different pointer types.

** I think it is undefined behavior, please feel free to correct my post if it is wrong.

Answer 4

Hope this code helps:

int main() {

    int arr[5] = {4,5,6,7,8};        
    int (*pa)[5] = &arr;
    int *pi = arr;

    for(int i = 0; i< 5; i++) {
        printf("\n%d %d", arr[i], (*pa)[i]);    
    }

    printf("\n0x%x -- 0x%x", pi, pa);
    pi++;
    pa++;
    printf("\n0x%x -- 0x%x", pi, pa);
}

prints the following:

4 4
5 5
6 6
7 7
8 8
0x5fb0be70 -- 0x5fb0be70
0x5fb0be74 -- 0x5fb0be84 

UPDATE: You can notice that pointer to integer incremented by 4 bytes (size of 32 bit integer) whereas pointer to array of integer incremented by 20 bytes (size of int arr[5] i.e. size of 5 int of 32 bit each). This demonstrates the difference.

Answer 5

Hope this code helps.

---

#include <stdio.h>
#include <stdlib.h>
#define MAXCOL 4
#define MAXROW 3

int main()
{      
      int i,j,k=1;
      int (*q)[MAXCOL];      //pointer to an array of integers

      /* As malloc is type casted to "int(*)[MAXCOL]" and every 
         element (as in *q) is 16 bytes long (I assume 4 bytes int), 
         in all 3*16=48 bytes will be allocated */

      q=(int(*)[MAXCOL])malloc(MAXROW*sizeof(*q)); 

      for(i=0; i<MAXROW; i++)
        for(j=0;j<MAXCOL;j++)
          q[i][j]=k++;


      for(i=0;i<MAXROW;i++){
        for(j=0;j<MAXCOL;j++)
          printf(" %2d ", q[i][j]);
        printf("\n");         
      } 
}

Answer 6

Who knows - this code exhibits undefined behavior:

printf("%d ",(*x)[i]);