C function declaration

can we declare a function in a header file in following way?

extern int ap( char[][] );

can w开发者_开发知识库e use char[][] in function?

No, you need to specify the last N-1 dimensions for an array.

extern int ap( char[][DIMENSION] );

For more information look here

For a two dimensional array, you have to supply a size for the last dimension otherwise the compiler doesn't know how to use it. (it's fine to omit the name though.)

like this:

extern int ap( char[][10] );

char[][] is not a valid array type because you cannot have arrays of an incomplete type, and char[] is incomplete. I know that's confusing because you really have two array types, so here's another example with the same problem: char a[3][]. The array a has length 3 and element type of char[], but char[] is, again, incomplete and this is invalid.

When you have a "multidimentional array", you really have an array of arrays. For example, typedef int U[3][5]; makes U an array of length 3 of arrays of length 5 of ints and is equivalent to typedef int H[5]; typedef H U[3];.

The reason you may omit the leftmost dimension with function parameters is because, with function parameters only, array types of the form T[N] are transformed into T*, and N can be left out, giving T[] to T*. This only applies at the "topmost" or "outermost" level.

So, all these function declarations are identical:

int f1(int a[3][5]);
int f2(int a[][5]);
int f3(int (*a)[5]);
typedef int T[5];
int f4(T a[3]);
int f5(T a[]);
int f6(T* a);

You can, of course, delete the parameter name a in any of the above declarations without changing them.

Yet, it is perfectly valid to omit parameter names in function declarations. When you define the function, however, you must give the array a name, and then you can refer to it by this name.

No, this is not allowed - it attempts to declare the parameter as a pointer to an incomplete array type.

The array type must be completed with a size, like this:

extern int ap( char[][10] );