Isn't double[][] equivalent to **double?

Question

I\'m asking this because my program have two functions to multiply matrices, they multiply only 4x4 and 4x1 matrices. The headers are:

 double** mult4x1(doub         


        

Answer 1

No... m1 is an array with four elements, each of which is an array of four elements. Same with m2. Even if the first "level" decays from an array into a pointer, the second "level" won't. The question is why? Let's look at some code:

/* double[4][4] implicitly converts to double(*)[4]
 * giving you a pointer to first element of m1 so
 * you get back an array of four doubles.
 */
double (*pm1a)[4] = m1[0];

/* This isn't right, but let's pretend that it was
 * and that what you got back was a pointer
 * to a pointer that pointed to m1 properly.
 */
double **pm1b = (double **)m1[0];

So, what would happen with our hypothetical code if we were to do pm1a[1] and pm1b[1]?

pm1a[1] is fine: it would correctly advance pm1a by 4 * sizeof(double) (since the pointer points to double[4]).

pm1b[1], on the other hand, would break: it would advance by the wrong amount: the size of a pointer to a double.

But this isn't all. There's a more subtle bug still. If both dimensions were to decay, the compiler could not know that an array is being accessed. It will happily interpret pm1b[1] as a pointer to a double. And then what happens? It will take whatever double value is stored at that location and treat it as a pointer to a double.

You can see why this would be a disaster.