二维数组地址和指向其第一个元素的对应指针

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 , ie char (*)[5] .

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