Thread: strings Vs. Char pointers

Originally Posted by Elysia

However, even so, you are accessing the array just like a 2D-pointer.

That's not really true: If a is defined as int a[10][10], a does not get passed into a function as a pointer to pointer, but a pointer to an array. (Array decay does not happen recursively, in other words.)

Originally Posted by Elysia

I don't know why the standard requires you to specify the inner/outmost dimension, whichever it is, when passing 2D-arrays.

I believe you, as breathtaking an admission as that is. Think for a minute: If you had a[10][10], so you needed to store 100 integers in memory somewhere, how would you arrange it so you could find a[0][1] and a[1][0] easily?

Right: you'd store things in rows, so that all the a[0][m]'s appear first, then all the a[1][m]'s next, and so on. In fact, you would know that the offset for a[x][y] would be 10*x+y (times the size of an int, of course). Now, if all you had was a bare pointer to the first element, how would you find a[1][0]?

Originally Posted by tabstop

Now, if all you had was a bare pointer to the first element, how would you find a[1][0]?

The same way that you would know how to find a[5] in a 1D array. You'd need to pass the size of the array to the function as well as the array. So if you pass a 2D array, I'm not sure why a pointer-to-pointer doesn't work since you could just pass 2 sizes along with the pointer?

Originally Posted by cpjust

The same way that you would know how to find a[5] in a 1D array. You'd need to pass the size of the array to the function as well as the array. So if you pass a 2D array, I'm not sure why a pointer-to-pointer doesn't work since you could just pass 2 sizes along with the pointer?

Which is why you need to pass a 2-D array as a[][10], just so that you can pass the size along with the pointer. If all you have is a pointer (which is all you get with int **), you don't have the size of the array there.

The point I'm (failing to) make here is that int ** and int[][10] are not the same type at all, since the first is "just a" pointer, while the second is a pointer plus enough information to make actually using the thing work.

Hmm, I've never thought of this stuff the way you guys(cas especially) are putting it.

So, if I'm getting this right:

If you take the address of a char * variable initialized as a string literal, you'd get the address of the pointer variable, but if you'd take the address of an array, you'd just get the address of the first element, which in fact means that there is no real variable 'storing' the address, so to speak?

And if my thoughts are on the right track, does someone care to explain how the sizeof macro works on arrays, when it doesn't on pointers?

I do understand the difference between the two more thoroughly now, but I still can't think of a way how the macro can work?