Question

I have this multi-dimensional array:

char marr[][3] = {{"abc"},{"def"}};

Now if we encounter the expression *marr by definition (ISO/IEC 9899:1999) it says (and I quote)

If the operand has type ‘pointer to type’, the result has type ‘type’

and we have in that expression that marr decays to a pointer to his first element which in this case is a pointer to an array so we get back ‘type’ array of size 3 when we we have the expression *marr. So my question is why when we do (*marr) + 1 we add 1 byte only to the address instead of 3 which is the size of the array.

Excuse my ignorance I am not a very bright person I get stuck sometimes on trivial things like this.

Thank you for your time.

Answer 1

It adds one because the type is char (1 byte). Just like:

char *p = 0x00;
++p; /* is now 0x01 */

When you dereference a char [][] it will be used as char * in an expression.

To add 3, you need to do the arithmetic first and then dereference:

*(marr+1)

You were doing:

(*marr)+1

which dereferences first.