Error using when using malloc for 2d array in C

Question

Why does this show a warning for dereferencing nullptr in C

Level1Tilemap->Map = (int*)((malloc(Level1Columns * sizeof(int))));

for (int i = 0; i < Level1Columns; i++)
{
    Level1Tilemap->Map[i] = malloc(Level1Rows * sizeof(int));

    for (int j = 0; j < Level1Rows; j++)
    {
        Level1Tilemap->Map[i][j] = Level1MapStaticArray[i][j];
    }
}

I am using malloc to create a 2D array of ints

But the editor shows warning and Level1Tilemap->Map has the memory address of nullptr

And the defination int** Map;

At a guess your editor is simply warning you that malloc can return a null pointer and you should check for that — Alan Birtles
– Alan Birtles, Commented Mar 15, 2022 at 11:38
In the first allocation, you use sizeof(int) but you probably want sizeof(int*) — ChrisMM
– ChrisMM, Commented Mar 15, 2022 at 11:39
Probably not your issue, but you want (int*)((malloc(Level1Columns * sizeof(int)))); -> (int*)((malloc(Level1Columns * sizeof(int*)))); — Mike Vine
– Mike Vine, Commented Mar 15, 2022 at 11:39
Check for malloc returning NULL, and if it does, use perror() to print it. — einpoklum
– einpoklum, Commented Mar 15, 2022 at 11:39
Is your Level1Tilemap properly set to something? And is it Level1Tilemap->Map = (int**)malloc... you want? — mmixLinus
– mmixLinus, Commented Mar 15, 2022 at 11:46

Lundin · Accepted Answer · 2022-03-15 11:44:08Z

2

You might have meant to use a pointer-to-pointer, in which case sizeof(int) should have been sizeof(int*). However, don't do that, it's obscure and slow, for no good reason.

Instead do

size_t size = sizeof(int[Level1Columns][Level1Rows]);

int (*something)[Level1Rows] = malloc(size);
if(something == NULL) { /* error handling */ }

memcpy(something, Level1MapStaticArray, size);
...
free(something);

Apart from reducing code complexity, this will significantly outperform the code you have currently.

answered Mar 15, 2022 at 11:44

Lundin

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Luminous · Accepted Answer · 2022-03-15 11:44:37Z

-1

// map is an array of (int*), and the length is column
int** map = (int**) malloc(column * sizeof(int*));

for (int i = 0; i < column; ++i) {
    // map[i] is an array of int, and the length is row
    map[i] = malloc(row * sizeof(int*));

    for (int j = 0; j < row; ++j)
        map[i][j] = 0;
}

answered Mar 15, 2022 at 11:44

Luminous

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2 Comments

Lundin Over a year ago

You fixed the first malloc call and broke the second one instead...

Lundin Over a year ago

Anyway, there is no sensible reason why you would ever want to use a fragmented malloc like this, when dealing with a plain 2D int array.

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