How would you declare a two dimensional array of pointers in C?

Question

...without using typedef.

My boss claims he was once asked this in an interview, and when he gave his answer the interviewers told him he couldn't use typedefs because it was poor style.

Regardless, he likes to throw the question out at people just to see if they can get it right , usually at new programmer lunches. No one ever gets it right (especially without a pen and paper or computer handy). I want to be ready next time he tries to stump someone with it >:D

Solution 1

void* array[m][n];

Will give you a 2D array of void pointers. I'm not sure what's so confusing about this, unless I misunderstood your question.

Solution 2

2D array of pointers to what?

T *p[N][M];     // N-element array of M-element array of pointer to T
T (*p[N][M])(); // N-element array of M-element array of pointer to 
                // function returning T

If we're talking about pointers to 2D arrays, then things only get slightly more interesting:

T a[N][M];            // N-element array of M-element array of T
T (*p)[M] = a;        // Pointer to M-element array of T
T (**p2)[M] = &p;     // Pointer to pointer to M-element array of T
T (*p3)[N][M] = &a;   // Pointer to N-element array of M-element 
                      // array of T
T (**p4)[N][M] = &p3; // Pointer to pointer to N-element array of 
                      // M-element array of T

Edit: wait, I think I may be getting what you're after.

T *(*a[N])[M];        // a is an N-element array of pointer to M-element
                      // array of pointer to T

Edit: Now we get really silly:

T *(*(*a)[N])[M];     // a is a pointer to an N-element array of 
                      // pointer to M-element array of pointer to T

T *(*(*(*f)())[N])[M];  // f is a pointer to a function returning
                        // a pointer to an N-element array of pointer
                        // to M-element array of pointer to T

T *(*(*f[N])())[M];     // f is an N-element array of pointer to 
                        // function returning pointer to M-element 
                        // array of pointer to T

And for the pathologically insane:

T *(*(*(*(*(*f)())[N])())[M])(); // f is a pointer to a function 
                                 // returning a pointer to a N-element
                                 // array of pointer to function 
                                 // returning M-element array of
                                 // pointer to function returning
                                 // pointer to T

Typedefs are for wusses.

Solution 3

void*** p2DArray = (void***)malloc( numAxis1 * sizeof( void** ) );

int count = 0;
while( count < numAxis1 )
{
    p2DArray[count] = (void**)malloc( numAxis2 * sizeof( void* ) );
    count++;
}

And there ya go. You can now access the array by going p2DArray[n][m] and get at the void* stored there.

Don't see why you'd need to use typedefs anyway ...

Edit: hahahaha or what avakar suggested ;)