动态分配2D数组

Question

My task is to make a 2D array 2xn from values given in A and B which have n elements.

I found this way to allocate memory for 2D array and it works but I don't understand if or why it is correct

int **tab2D(int A[],int B[],int n)
{
    int **newTab = malloc(n*sizeof(int*));
    newTab[0] = A;
    newTab[1] = B;
    return newTab;
}

I know there are other ways to do this but I'm curious about this one.

Answer 1

As I understand your code it should work like this.

int **tab2D(int A[],int B[],int n)

can be seen as

int **tab2D(int *A,int *B, int n)

so you pass pointers to the two arrays which are already allocated. You then allocate some memory for the pointers-to-pointers

 int **newTab = malloc(n*sizeof(int*));

which I think should be

 int **newTab = malloc(2*sizeof(int*));

instead, because you have A and B , which are both the same length n , I assume. Then you dereference the new pointer-to-pointers and assign the pointers to your arrays

newTab[0] = A;
newTab[1] = B;

which could be written as

*newTab = A;
*(newTab + 1) = B;

Answer 2

This is not a 2D array of int s, is an array of pointers to int .

It works because each element of the array points to an address containing an array of int s

A similar code using the stack, just to show why it works:

int a[] = {1, 2};
int b[] = {3, 4};
int *arr[] = {a, b};

A real 2D array is supposed to work in contiguous areas (without fragmentation), you can use pointers to VLA's to achieve this:

int rows, cols;

scanf("%d %d", &rows, &cols);

int (*arr)[cols] = malloc(sizeof(int [rows][cols]));