Two dimensional arrays and pointers

Question

I have the following code snippet:

char board[3][3] = {
                     {'1','2','3'},
                     {'4','5','6'},
                     {'7','8','9'}
                   };

printf("address of board : %p\n", &board);
printf("address of board[0] : %p\n", &board[0]);

Both printf() statements all print the same value: 0x0013ff67

1. As per my knowledge, board (i.e) array name represents the address of the first subarray (i.e) board[0] and

2. board[0] represents the address of first element in the first array (i.e) board[0][0]

Why am I getting the same address in all my printf() statements? I expect different addresses for both statements.

I am pretty new to this stuff and don't understand this behavior. Kindly enlighten me.

Answer 1

As per my knowledge, board (i.e) array name represents the address of the first subarray (i.e) board[0]

This is only true if board is used outside of these contexts

• As operand of the & operator
• As operand of sizeof

When any of that applies, expression board represents the array and keeps having the type of the array (char[3][3]). Applying the & operator to it results in getting the address of the array, which of course equals the address of its first element, merely having a different type (char(*)[3][3] instead of char(*)[3]). The same that is true about the array board is true about its first sub array board[0].

When you use it outside of those contexts, you get the address of the first element (subarray in your case). That address is not an object but just a value. Value have no address, but objects have. Trying to apply & on it would fail. For example

// error: trying to apply '&' on something that has no address
&(1 ? board : board)

Note that anything said above applies to C; not necessarily to C++.

Answer 2

Though it's a 2D array, inside the memory it will be represented as linear array. so when you say, board[0][0] it still points to the first element in that linear array and hence the same memory address.

Answer 3

It may be the case that you know an object-oriented language such as Java or Python, and now you are learning the C language. The difference between Java and C when thinking about char board[3][3] is that in C the board variable is represented in memory as 9 characters at adjacent memory addresses. Like so:

board: 1 2 3 4 5 6 7 8 9

In C, &board yields the same memory address as &board[0] and &board[0][0].

In contrast to this, in Java the variable would be declared as char[][] board and its memory representation would conceptually look like this:

board: ptr(A) ptr(B) ptr(C)
A:     1 2 3
B:     4 5 6
C:     7 8 9

where ptr(x) points to memory address of x. So, in Java, board points to a different memory address than board[0].

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You say In C, &board yields the same memory address as &board[0] and &board[0][0]. But i am able to access the first element only via board[0][0] (or) *board[0] (or) **board. Why is it so??

Although the expressions &board and &board[0] and &board[0][0] yield the same address, the type system of the C language is preventing you from accessing the char value. In a C compiler, the types are (conceptually):

board:       type char[3][3]
board[0]:    type char[3]
board[0][0]: type char

Assuming a variable of type char, we can write:

char c;
c = board[0][0];

but cannot write:

char c;
c = board;    // Error
c = board[0]; // Error

because the type on the left side is incompatible with the type on the right side of the assignment.

If you are sure that an address points to a char, you can use a type cast:

char c;
c = *(char*)board;    // Works OK
c = *(char*)board[0]; // Works OK

The downside is that such type casts may lead to coding bugs.

Answer 4

Of course this will print the same address.
Think about 2D arrays like this for a minute,

int **ptr;

The address *ptr does equal &(**ptr).

Because basically, that's what your code is doing.

and remember that in memory, 2D arrays will be mapped linearly.