How does we perform pointer-arithmetic on multidimensional arrays in C? Kudos if you also explain how multidimensional arrays are actually implemented in C.

Question

anonymous · Answer

int i, j;
int a[][3] = {{4,1,2}, {3,6,5}, {9,9,5}};
/* Prints all of column 1*/
for (j = 0; j < 3; ++j) {
    fprintf(stdout, "%d
", a[0][j]);
}

what if I want to replace a[0][j] with a pointer-arithmetic expression?

hero · Answer

?

anonymous · Answer

You literally just replace it with pointer arithmetic ... I don't see where your problem is.

anonymous · Answer

That's just it: I need an example of how it is done in my particular case up there

anonymous · Answer

Since multidimensional arrays are just arrays of arrays, I guess a[0][j] can be replaced by *(*(a + 0) + j);

*(a + 0) gets out an array (the first ), and  the remainder of the expression gets out the  element at index j of that array.

actually, now that I've explained it myself, I think that comment is a lie; it prints out all elements of the first row of that matrix :-P

anonymous · Answer

simply switch the 0 and the j, and it correctly prints out the 'column'

anonymous · Answer

...I feel enlightened.

anonymous · Answer

A way to think about it is that in C there isn't really such a thing as an array as implemented in other languages, as an array is simply a pointer. Hope that helps.

anonymous · Answer

right; a is effectively a int**

anonymous · Answer

or am i mistaken?

anonymous · Answer

Yes, for example, a 2d array simply points to the beginning of a 1d array of pointers.

anonymous · Answer

I think I'm mistaken.