C variable declaration in gcc compiler - compile time error

Assume the following C variable declaration:

int *A[10], B[10][10];

Of the following expressions:

1. A[2]
2. A[2][3]
3. B[1]
4. B[2][3]

Which will not give compile time errors if used as left hand sides 开发者_如何转开发of assignment statements in a C program.

A) 1, 2 and 4 only

B) 2, 3 and 4 only

C) 2 and 4 only

D) 4 only

I have tried this on a gcc compiler. I assigned the value '0' to all the above variables. Only the third one showed an error. I can't really understand the reason. Can someone please explain the reason?

1. You can assign 0 to A[2], because A is an array of pointers, and you can assign 0 to a pointer (it's a NULL pointer).
2. You can assign 0 to A[2][3], because at this level you're working with the int.
3. You cannot assign 0 to B[1], because B is an array of arrays, and 0 is a scalar.
4. See 2.

Break the declaration into:

int *A[10];
int B[10][10];

You can see that A[10] is really an array of pointers, while B[10][10] is an array of integer arrays. The reason why you cannot assign an integer to B[1] is because B[1] is supposed to be of type int[] (an array), and you can't overwrite it with an int value.

Assigning to A[2] works because you're just pointing that array element to some other value, in this case an int.

1. A[2] = 0 is correct, because A[2] = NULL is obviously correct and NULL is defined as 0. However if you try some value other than 0, you should encounter some type transition error.
2. A[2][3] = 0 is correct, in terms of syntax, as this statement is understood by the compiler as * (* (A + 2) + 3) = 0; Or more clearly, consider it as "int* p = A[2]; *(p + 3) = 0";
3. B[1] is incorrect; the superficial reason is B[1] points to an array and you cannot assign an array; the essential reason is that B[1] is translated into some address value by the compiler, so it cannot work as a left value. Or one can consider it this way: There is no corresponding memory cell for B[1].
4. B[2][3] = 0 because this is how it works :)