C Array of const unsigned char[][]

Question

I'm a little turned around on the exact syntax for doing this. I have one array defined as such:

const unsigned char ARR[SIZE][SIZE] = {...}

and I want to have it in an array so I can do something to the effect of

ARR2[0] = ARR

I've tried const unsigned char ARR2[][SIZE][SIZE] = {ARR} and const unsigned char* ARR2[SIZE][SIZE] = {ARR} , but neither of those worked. Can someone point out the correct syntax for having a constant array of constant two dimensional arrays of unsigned characters?

Answer 1

From your comment to haccks' answer, it sounds like you want this:

const unsigned char ARR[SIZE][SIZE] = {...};
const unsigned char (*ARR2[])[SIZE][SIZE] = {&ARR};

If you want ARR2 itself to be const , that would be like this:

const unsigned char (* const ARR2[])[SIZE][SIZE] = {&ARR};

It should perhaps also be noted that you cannot use this abomination to access ARR[x][y] as ARR2[0][x][y] , but need to do (*ARR2[0])[x][y] (or, equivalently, ARR2[0][0][x][y] ).

Another corollary that might be noteworthy is that you cannot assign ARR2 with ARR2[0] = ARR , but you need explicitly to do ARR2[0] = &ARR (at least to avoid warnings). The reason is that ARR degenerates to a pointer to its first element , that is, a const unsigned char * whose value is &ARR[0][0] , whereas ARR2[0] expects to be assigned a const unsigned char (*)[SIZE][SIZE] , which is obtained with &ARR . While the pointer value is identical in both cases, the types aren't.

Answer 2

Seems like what you are trying to achieve is impossible since left operand of assignment operator can be only of arithmetic, structure / union or pointer type, but not an array one. See here for details: 6.5.16.1 Simple assignment .

So the only way for you is only memcpy() 'ing.

Answer 3

Try this

const unsigned char (* ARR2)[SIZE] = ARR; 

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You mentioned in your comment that you want ARR2 to be an array that can hold array ARR as it's first element but, this is not possible . ARR is an array and it will converted to pointer to it's first element expect when it's an operand of unary & and sizeof operator.