How does a C++ program allocate memory for an array?(NOT DYNAMICALLY)

Question

When we create an array, for example int arr[2][4];, its size becomes 2x4x4 = 32, but if we do this dynamically, its size becomes more then 32.

For instance, I wrote this code:

#include <iostream>
#include <stdlib.h>

using namespace std;
// arr[4][2]
int main(){
    int **arr;
    arr = (int**)malloc(4*sizeof(int*));
    
    int i;
    for(i=0;i<4;i++){
        *(arr+i) = (int*)malloc(2*sizeof(int));
    }
    arr[3][1] = 4;
    cout << arr[3][1];


    return 0;
}

There are 8 + 4x8 + 4x2x4 = 72 bytes allocated. That's why I thought the program is doing different and I wonder how it does this.

Because, in a true array, there are no lookups for each row: all the elements are contiguous. — Weather Vane
– Weather Vane, Commented May 26, 2021 at 23:03
What do you mean "not dynamic"? Your code does dynamic allocation. A static array definition just reserves a contogous array of data elements at it's scope (local, or global), why would you expact these sizes to be equal ? — πάντα ῥεῖ
– πάντα ῥεῖ, Commented May 26, 2021 at 23:07
@πάνταῥεῖ I think the title means he's wondering about int arr[2][4]; — Barmar
– Barmar, Commented May 26, 2021 at 23:09
Yes, but because the elements are contiguous, the compiler needs only the base address to compute the location of any element. But when you allocated the array of pointers, the allocation for each row has no proximity to the other rows, and so a separate pointer was used for each row. — Weather Vane
– Weather Vane, Commented May 26, 2021 at 23:42

dbush · Accepted Answer · 2021-05-26 23:03:17Z

3

What you've done here isn't exactly a 2D array although it behaves like one syntactically. What you instead have is an array of pointers, and each of those pointers points to an array of int.

If you want to allocate an actually 2D array you would do so like this:

int (*arr)[2] = malloc(4 * sizeof *arr);

answered May 26, 2021 at 23:03

dbush

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1 Comment

Ahmet Talha Yıldırım Over a year ago

What if I want to allocate a 3D array?

Barmar · Accepted Answer · 2021-05-26 23:18:39Z

1

int arr[4][2] allocates a single block of memory, with indexes laid out like this:

0,0 0,1
1,0 1,1
2,0 2,1
3,0 3,1

Each element is 4 bytes to hold an int.

In your dynamic allocation method, you have 8 bytes for the arr variable itself. Then the first malloc() allocates an array like this:

ptr0 ptr1 ptr2 ptr3

This is 4x8 bytes.

Then you call malloc(2 * sizeof(int)) 4 times, which allocates 4x2x4 bytes.

answered May 26, 2021 at 23:18

Barmar

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4 Comments

Ahmet Talha Yıldırım Over a year ago

The first thing you said "int arr[4][2] allocates a single block of memory" is okay but then how arr behaves like a pointer of pointer. If it behaves like a pointer of pointer, I should be able to do it dynamically.

Barmar Over a year ago

If an array is used in a context where a pointer is needed, its value decays to a pointer to its first element.

Barmar Over a year ago

So if you write arr+1, it acts like an int[2]* pointer. Or if you pass it as a function argument.

Andreas Wenzel Over a year ago

@AhmetTalhaYıldırım: You may want to read this: What is array to pointer decay?

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