Two dimensional array address and corresponding pointer to its 1st element

Question

In terms of one dimensional array, its array name is also the address of the first element. So it is fine to assign it to a pointer, like below:

char data[5];
char* p_data=data;

So I think it should be the same with two dimensional array. The array name should be the address of the first element's address. So, I'd like to do something like this:

char data[5][5];
char** pp_data=data;

Then I get a warning saying the pointer type char** is incompatible with char[ ][ ].

Why does this happen? Do I comprehend the pointer and array concept wrong?

Answer 1

You're right that an array is often referred to by a pointer to its first element. But when you have the "two dimensional" array

char data[5][5];

what you actually have is an array of arrays. The first element of the array data is an array of 5 characters. So this code would work:

char (*pa_data)[5] = data;

Here pa_data is a pointer to array. The compiler won't complain about it, but it may or may not actually be useful to you.

It's true that a pointer-to-pointer like your char **pp_data can be made to act like a two-dimensional array, but you have to do some memory allocation for it to work. It turns out that in the array-of-arrays char data[5][5] there's no pointer-to-char for pp_data to be a pointer to. (In particular, you could not say something like pp_data = &data[0][0].)

See also this question in the C FAQ list.

Answer 2

Two dimensional array is actually an array of arrays. It means the first element of that array is an array. Therefore a two dimensional array will be converted to pointer to an array (its first element).

In

char data[5][5];  

when used in expression, wit some exception, data will be converted to pointer to its first element data[0].data[0] is an array of char. Therefore the type of data will become pointer to an array of 5 char, i.e. char (*)[5].

char ** and char (*)[5] are of different type, i.e. incompatible type.