Question

I have a function which tries to loop through a 2D array in a struct:

typedef struct node
{
    int grid[3][3];
} Node;


void someFunction(Node *node) {
     int grid[3][3] = node->grid;
     //loop through
}

When I try to compile this however I get a

mp.c:42: error: invalid initializer

Answer 1

You cannot assign arrays in C. It is simply not allowed. When you wrote:

int grid[3][3] = node->grid;

you were attempting to initialize the local array, grid, from the passed in node. If that were permitted (which it isn’t), then you’d not need a loop afterwards.

You can assign structures, though, even if they contain arrays, so if the local structure were a Node, you could have written:

Node local = *node;

You would not need to loop through the arrays afterwards to initialize local.

You can loop through the arrays, doing the copy one element at a time:

for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
        grid[i][j] = node->grid[i][j];

You can also use memmove() or memcpy():

int grid[3][3];

assert(sizeof(grid) == sizeof(node->grid));
memcpy(grid, node->grid, sizeof(grid));

---

At one time, another answer suggested:

Change the line:

int grid[3][3] = node->grid;

to:

int **grid = node->grid;

I noted that this will not work – and was legitimately challenged to explain why. That requires space and formatting.

First of all, the compiler notes:

warning: initialization from incompatible pointer type

That is saying ‘you are playing with fire’.

Let’s suppose we ignore that warning. The local grid now points at the top-left corner of the array (if you see arrays growing down and across from left-to-right). The value stored there is a plain number, not an initialized pointer, but when the compiler evaluates grid[0], it is forced to assume that will produce a pointer. If node->grid[0][0] contains a zero, you will probably get a segmentation fault and core dump for dereferencing a null pointer (on the assumption that pointers and int are the same size, which is generally true on 32-bit systems), or some other undefined behaviour. If node->grid[0][0] contains another value, then the behaviour is still undefined, but not quite as predictable.