Initializing 2d array in C using memset

Question

I am getting an exception when printing a two-dimensional array after initializing it using memset. If I initialize each element and then access it, then there is no error. Please help to figure out the bug in the code.

#include "stdafx.h"
#include "stdlib.h"
int r = 0;
int c = 0;
int **twoDArray() {

    int *arr[5];

    for (r = 0; r < 5; r++) {

        arr[r] = (int*)malloc(2 * sizeof(int));

    }
   // 
    for (r = 0; r < 5; r++) {
        for (c = 0; c < 2; c++) {
            arr[r][c] = 1;
            printf("%d \n", arr[r][c]);
        }
    }
    memset(arr, 0, 2 * 5 * sizeof(arr[0][0]));
    for (r = 0; r < 5; r++) {
        for (c = 0; c < 2; c++) {
            printf("%d \n", arr[r][c]); //getting exception here.
        }
    }
    return arr;

}


int main()
{
    int **arr;
    arr = twoDArray();
    for (r = 0; r < 5; r++) {
        for (c = 0; c < 2; c++) {
            arr[r][c] = 0;
            printf("%d \n", arr[r][c]);
        }
    }
    return 0;
}

Your program is broken regardless of whether you use memset or not. Your twoDArray function returns a pointer to a local array arr. The latter is automatically destroyed when the function exits. Trying to use the returned result in main triggers undefined behavior. — AnT stands with Russia
– AnT stands with Russia, Commented May 26, 2018 at 17:53

Angew is no longer proud of SO · Accepted Answer · 2018-05-26 17:50:31Z

You don't have a 2D array. You have a 1D array of pointers to 1D arrays; each of these arrays can be (and 99.9% likely is) at a different address in memory. You therefore cannot treat them as one continuous block, as memset does.

The best (and most cache-friendly) option is to allocate the entire array dynamically:

int *arr = malloc(2 * 5 * sizeof(int));

This will require you to then do manual indexing:

for (r = 0; r < 5; r++) {
    for (c = 0; c < 2; c++) {
        arr[r * 2 + c] = 1;
        printf("%d \n", arr[r * 2 + c]);
    }
}

If you know the array size at compile time (if the 2 and 5 are actual constants), you can also make arr a pointer to an actual array, to allow syntactic 2D indexing:

typedef int Array[2];

Array *arr = malloc(5 * sizeof(Array));

for (r = 0; r < 5; r++) {
    for (c = 0; c < 2; c++) {
        arr[r][c] = 1;
        printf("%d \n", arr[r][c]);
    }
}

Note that in your original code, there was also the problem you were returning a pointer into arr, which is a local variable, making the pointer immediately dangling.

user2736738 · Accepted Answer · 2018-05-26 17:59:38Z

3

This is an array of pointers. You can be sure of two things

arr[0] to arr[4] is contiguous.
The chunk allocated is contiguous.

But who told you that the memory block allocated by each malloc which each of the pointer points to is contiguous also? Nobody did. You assumed it wrong.

In your case that's definitely not the case. That's why you invoked undefined behavior by accessing memory out of bound. (in memset).

Do this though - it would work

int (*arr)[2];
arr = malloc(sizeof *arr * 5);
// do memset - it's fine.

Don't cast the return value of malloc. It's redundant.

Something like this in earlier case:-

                                  +-------------+
        +-------+         +------>+  arr[0][0]  |
arr[0]  |       |         |       |      |    arr[0][1]
        |   +-------------+       +------+------+
        +-------+
        |       |     +------+------+
arr[1]  |    +------->+      |      |
        +-------+     |      |      |
        |       |     +-------------+              +------+------+
arr[2]  |    +-------------------------------------+      |      |
        +-------+                                  |      |      |
        |     | |                 +------+------+  +------+------+
arr[3]  |     +------------------>+      |      |
        +-------+                 |      |      |
        |       |                 +------+------+
arr[4]  |    +  |
        +-------+
             |                        +------+------+
             +----------------------->+      |      |
                                      |      |      |
                                      +------+------+

edited May 26, 2018 at 17:59

answered May 26, 2018 at 17:44

user2736738

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3 Comments

Soner from The Ottoman Empire Over a year ago

#rez to myself for future.

user2736738 Over a year ago

@snr>; I am not sure what you mean...#rez?

Soner from The Ottoman Empire Over a year ago

I'm impressed your answer. #rez means I leave my comment here to glance at the answer in future.

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