pointers - malloc in C, but use multi-dimensional array syntax

Question

Is there any way to malloc a large array, but refer to it with 2D syntax? I want something like:

int *memory = (int *)malloc(sizeof(int)*400*200);
int MAGICVAR = ...;
MAGICVAR[20][10] = 3; //sets the (200*20 + 10)th element

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UPDATE: This was important to mention: I just want to have one contiguous block of memory. I just don't want to write a macro like:

#define INDX(a,b) (a*200+b);

and then refer to my blob like:

memory[INDX(a,b)];

I'd much prefer:

memory[a][b];

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UPDATE: I understand the compiler has no way of knowing as-is. I'd be willing to supply extra information, something like:

int *MAGICVAR[][200] = memory;

Does no syntax like this exist? Note the reason I don't just use a fixed width array is that it is too big to place on the stack.

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UPDATE: OK guys, I can do this:

void toldyou(char MAGICVAR[][286][5]) {
  //use MAGICVAR
}

//from another function:
  char *memory = (char *)malloc(sizeof(char)*1820*286*5);
  fool(memory);

I get a warning, passing arg 1 of toldyou from incompatible pointer type, but the code works, and I've verified that the same locations are accessed. Is there any way to do this without using another function?

Answer 1

Yes, you can do this, and no, you don't need another array of pointers like most of the other answers are telling you. The invocation you want is just:

int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
MAGICVAR[20][10] = 3; // sets the (200*20 + 10)th element

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If you wish to declare a function returning such a pointer, you can either do it like this:

int (*func(void))[200]
{
    int (*MAGICVAR)[200] = malloc(400 * sizeof *MAGICVAR);
    MAGICVAR[20][10] = 3;

    return MAGICVAR;
}

Or use a typedef, which makes it a bit clearer:

typedef int (*arrayptr)[200];

arrayptr function(void)
{
    /* ... */