Pointer to array type confusion

Question

I thought that if I create a pointer to array type, I will be able to print all of its elements. I wrote a program like this,

#include <stdio.h>

int main()
{


    int arr[] = {4,2,3};
    //here arr is a (*arr)[3] type
    int (*p)[3], i;
    p = &arr;
    for(i = 0 ; i < 3 ; i++)
            printf("%d ", *(p)[i]);
    printf("\n");
    return 0;
}

output : 4 0 -673566907

My understanding is p will point to an array of 3 elements. p+1 will point to another array of 3 elements.

arr=x   [ 1      2     3 ]
 y       [x]  [x+1] [x+2]

p=y(address of arr)
*p=x
**p = 1
*(*p+1) = 2
*(*p+2) = 3

I can print array like above. Does it mean p is actually double pointer?

P.S Is it correct way to read

*(p)[1] should be read as "p is an array of 1 pointer to"
(*p)[1] should be read as "p is a pointer to an array of 1 element"
when we print

printf("%d ", (*p)[1]); //here it should be read as p is a pointer to the second element"

Chris Turner · Accepted Answer · 2017-03-23 11:15:54Z

2

You're dereferencing your pointer p in the printf statement incorrectly. Your code is acting as if p is an array of pointers to int instead of a pointer to an array of int. Instead of

printf("%d ", *(p)[i])

you want

printf("%d ", (*p)[i])

because you have to first turn p into the int[3] and then index into it.

answered Mar 23, 2017 at 11:15

Chris Turner

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Comments

clickMe · Accepted Answer · 2017-03-23 11:28:40Z

1

Change the Line

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

to

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

It will access the pointer-to-array first, due to referencing (*p), and then iterate through the content of the array with the []-operator.

Consider the second iteration in the for-loop from your code which equals to:

printf("%d ", *(p + 1));

Trying to access the memory location p+1 with the * leads to undefined behavior as others have already pointed out.

edited Mar 23, 2017 at 11:28

answered Mar 23, 2017 at 11:13

clickMe

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4 Comments

Sourav Ghosh Over a year ago

My understanding is p will point to an array of 3 elements. p+1 will point to another array of 3 elements....so it does not quite match your statement here.

clickMe Over a year ago

@acidlategamer yes. the parenthese is redundant in this case

acidlategamer Over a year ago

While printing if I am using (*p)[1], then I am pointing to address of the second element or value, or getting only the address of the second value?

clickMe Over a year ago

The expression (*p) gives you the memory address where the array arr is located. You then apply the [ ]operator to it. Thus, (*p)[1] is equal to arr[1].

acidlategamer · Accepted Answer · 2017-03-23 11:42:53Z

1

I will post my understanding from this thread,

pointer to array type

int arr[4] = {2,4,1,7};
int (*p)[4];//here p is a pointer, pointing to an array of 4 elements
p = &arr;

now if i want to access elements of 0th array

*p = *(p+0); //first element address pointed by p

*p+1 = *(p+0)+1;//second element addres pointed by p

and so on

now

**p = *(*(p+0)) ;//dreference of the 0th element of 0th array. equals to (*p)[0] expression.

*(*p+1) = *((*p+0)+1);//dereference of the 1st element of 0th array. equals to (*p)[1] expression

*(p)[1] = *(p+0)[1] =*(p+(1))= *(p+1) = next array.

answered Mar 23, 2017 at 11:42

acidlategamer

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Comments

Sourav Ghosh · Accepted Answer · 2017-03-23 11:12:15Z

0

No, for any value other than 0 for i

for(i = 0 ; i < 3 ; i++)
        printf("%d ", *(p)[i]);

causes undefined behavior as you try to access invalid memory.

Therefore, the is no scope of p being a pointer-to-pointer, it is a pointer-to-array, and as they say, arrays are not pointers and vice-versa.

You can, however, take a pointer to the starting of the array and then loop over till you have valid elements in the array, that will be allowed.

answered Mar 23, 2017 at 11:12

Sourav Ghosh

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Comments

Fatemeh Karimi · Accepted Answer · 2017-03-23 17:33:12Z

0

I didn't understand your code well, but I edited it in this way and it worked. hope it can help you.

#include <stdio.h>
int main()
{
    int arr[] = {4, 2, 3};
    int *p[3];
    int i = 0;
    p[0] = arr;

    for (int i = 0; i < 3; i ++)
        printf("%d\n",*(p[0] + i) );

    return 0;
}

answered Mar 23, 2017 at 17:33

Fatemeh Karimi

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