How to access 1D array as 2D array?

Question

I'm trying to access 1D array as 2D array. But it falls into Segv. Below is snippet the i had written. Can anyone please take a look into this?

void printarray(int **a){


    printf("#####2D access... \n");
    for(int i=0;i<2;i++){
        for(int j=0;j<2;j++){

            printf("## %u-->%d \n", &a[i][j],a[i][j]);
        }
        }

}

int main(){
    int a[4] = {10,20,30,40};
    printf("%u %u %u \n", &a, a, &a[0]);
    printf("%u %u %u %u \n", &a[0], &a[1], &a[2], &a[3], &a[4]);
    printarray((int **)a);
    return 0;
}

And vice-versa(Accessing 2D array as 1D array) situation is simply handled by type casting.Below is the snippet and this works fine.

void printarray(int *a){
    printf("#####1D access... \n");
    for(int i=0;i<4;i++){
        printf("## %u-->%d \n", &a[i],a[i]);
    }
}

int main(){
    int a[2][2] = {
        {10,20},{30,40}
    };
    printf("%u %u %u %u \n", &a, a, a[0], &a[0]);
    printf("%u %u \n", a[0], &a[0]);
    printf("%u %u \n", a[1], &a[1]);

    printarray((int *)a);
    return 0;
}

thanks, Hari

in your first example, which dimensions of a 2D-array do you expect when having a 1D-array of size 4; shall it be 1x4,2x2, 4x1? The compiler won't guess :-) — Stephan Lechner
– Stephan Lechner, Commented Jul 27, 2018 at 13:16
@StephanLechner ultimately this 2D array(1x4 or 2x2 or 4x1) are stored sequentially, then we can access the any element using the base address.This sown In the second example by typecasting. Similarly in 1st case also why can't i access elements in 2D manner for the 1D array ? — Hari Krishna Nalla
– Hari Krishna Nalla, Commented Jul 27, 2018 at 13:23
a** is an pointer to a pointer. Not a two dimensional array. a[i][j] is probably trying to use a[i] as a pointer and then getting thejth element after that pointer. It is (a[i])[j] so a[0][1] is going to be ((int*)(10))[1]; on a 32 bit system. — William J Bagshaw
– William J Bagshaw, Commented Jul 27, 2018 at 13:24
@WilliamJBagshaw It something interesting. i didn't get you. Can you please explain this line a[0][1]-->((int*)(10))[1]. My understanding is that a[0][1]-->*((baseaddrof(a)+0)+1). In this how can *(baseaddrof(a)+0) this line might refer to (int)(10) ? — Hari Krishna Nalla
– Hari Krishna Nalla, Commented Jul 27, 2018 at 13:36

Killzone Kid · Accepted Answer · 2018-07-27 13:42:42Z

2

You want to use reinterpret_cast to get a reference to the data stored in your array:

#include <iostream>

void printarray(int a[2][2])
{
    printf("#####2D access... \n");
    for (int i = 0; i<2; i++)
    {
        for (int j = 0; j<2; j++)
        {

            printf("## %p-->%d \n", &a[i][j], a[i][j]);
        }
    }
}

void printarray(int b[4])
{
    printf("#####1D access... \n");
    for (int i = 0; i<4; i++)
    {
        printf("## %p-->%d \n", &b[i], b[i]);
    }
}

int main()
{
    int a[4] = { 10,20,30,40 };
    int(&arr)[2][2] = reinterpret_cast<int(&)[2][2]>(a);
    printarray(arr);

    int b[2][2] = {{ 10,20 },{ 30,40 }};
    int(&brr)[4] = reinterpret_cast<int(&)[4]>(b);
    printarray(brr);

    return 0;
}

Example: https://ideone.com/thc55R

#####2D access... 
## 0x7fff708c4a90-->10 
## 0x7fff708c4a94-->20 
## 0x7fff708c4a98-->30 
## 0x7fff708c4a9c-->40 
#####1D access... 
## 0x7fff708c4aa0-->10 
## 0x7fff708c4aa4-->20 
## 0x7fff708c4aa8-->30 
## 0x7fff708c4aac-->40

answered Jul 27, 2018 at 13:42

Killzone Kid

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6 Comments

Hari Krishna Nalla Over a year ago

thanks for your answer..and I heard that reinterpret cast is dangerous..so any idea to use any other method.?

Killzone Kid Over a year ago

@nhkrishna Yes, avoid raw arrays all together, use std::array or std::vector instead

Hari Krishna Nalla Over a year ago

I have used one dimensional dynamic array, then 2D accessing output gives different results. Here is the code ideone.com/oWoIAq . Can you please take look into this ?

Killzone Kid Over a year ago

@nhkrishna a is a pointer and not the array. If you dereference it reinterpret_cast<int(&)[2][2]>(*a); it might work but would probably result in undefined behaviour.

Hari Krishna Nalla Over a year ago

yes, if i dereference the pointer and use the reinterpret_cast then it works perfectly fine. Thank you :-)

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