Thread: Multidimensional arrays

Originally Posted by Sebastiani

Code:

	printf( "address of data: %x \n", ( size_t* )data );
	printf( "address of data + 1: %x \n", ( size_t* )( data + 1 ) );

%p is the correct format for pointers, and it expects a void pointer. Alternatively, cast to unsigned, and continue to use %x.

Using the wrong format specifiers for printf is undefined behavior.

Originally Posted by Niels_M

That is what I meant by "accessing" the array. But OK, I guess this is just how it is with 2D-arrays.

I just thought there was a logical explanation like there is with string literals, i.e. that in

Code:

*test[] = {"hi", "mi"}

, then test[0] points to the first character in the string literal "hi" etc..

I think the difference here (if there is a difference here, which I'm not entirely convinced thiere is) is that in your test array the strings hi and mi live god knows where -- the values in the array are pointers to somewhere else, while in a 2D array there are no pointers involved anywhere (excepting the conversion of the name of the array to a pointer when used by itself). So your test array might store 0x403000 and 0x403003, while if you did

Code:

char test2[2][3] = {"hi", "mi"};

your test2 array will store 'h', 'i', 0, 'm', 'i', 0 -- no pointers anywhere. (Now test2[1], by itself, is the name of an array, so it will be converted to the address of the beginning of the second "row" of your array.)

EDIT: Or perhaps you're misunderstanding "second element". test2[1] is the second element of the test2 array -- which is to say, it is the second row/string/whatever you want to call it. It is not the second character stored, which is the second element of the first element of the array.

Originally Posted by Niels_M

I just thought there was a logical explanation like there is with string literals, i.e. that in

Code:

*test[] = {"hi", "mi"}

, then test[0] points to the first character in the string literal "hi" etc..

The same is true for 2D arrays; asd[0] points to the first integer of the array asd[0]

Code:

asd[0] = &asd[0][0]; /* points to the number 1 */

Originally Posted by tabstop

(Now test2[1], by itself, is the name of an array, so it will be converted to the address of the beginning of the second "row" of your array.)

Ok, I think I understand it now. We have

Code:

char testarr[][25] ) {"thanks", "everybody"}

Here (testarr+1) is the address of "everybody", and thus *(testarr+1) is the "name" of the array, and thus the address of the first element. Thus *(*(testarr+1)+1) gives us a 'v'.