I have a void* I am getting in some function which is actually a two-dimensional int array. I want to send it as an argument to a function that expects a two dimensional array. What is the BEST way to cast it properly?
void foo(void* val){ //How to cast val in order to send to bar?? bar() } void bar(int val[2][2]){ //Do something } 
bar((int(*)[2]) val); (As Carl Norum states, the cast isn't even actually required; but it has the advantage of giving you a compiler warning if you accidentally pass it to a function expecting, say, a int(*)[3].)
void *, when I can, since the round-trip is particularly error-prone. One alternative to using a multidimensional array is to wrap up the int[2] into a struct. Another is to use a jagged array, though obviously that means having to deal with memory allocation.
int arr1[2][2]; int arr2[2][2] = arr1; doesn't work. When an array is a parameter to a function, it's essentially just set up as a pointer to whatever is passed in; but when you declare an array as a local variable inside a function, the space for it is allocated on the stack, so it's not just a matter of reinterpreting an existing pointer.p is a void * that you know points to an int[...][2], you can write int (*arr)[2] = p; to declare an arr that has the same underlying data, but has type int[...][2]. You can then refer to arr[0][0], arr[0][1], arr[1][0], and arr[1][1], just as you'd expect. You cannot cast it to an int**, because in C, a multidimensional array is not an array of pointers. Internally it's basically stored as a one-dimensional array, and the compiler does all the work to treat arr[i][j] as ((int *)p)[2*i+j] (2 being the number of elements per row).
void *translate itself automatically in C?