Question

Possible Duplicate:
Confusion over C++ pointer and reference topic

Suppose I am passing

int arr[10];

as a parameter to a function.

Are all of these valid function prototypes? How do they differ in terms of arguments and why?

This is what I know so far (not sure if correct or not)

1. void foo(int &arr);      //a reference to an array, which preserve the size attribute?
2. void foo(int &arr[]);    //the same (is it)?
3. void foo(int (&arr)[]);  //I don't know
4. void foo(int &arr[10]);  //is it the same as 2?
5. void foo(int (&arr)[10]);//no idea

6. void foo(int *arr);      //a decayed pointer of the array, pointing to the first element of the array, losing the size attribute?

7. void foo(int *arr[]);    //the same (is it?)
8. void foo(int (*arr)[]);  //a pointer to an array, preserve the size
9. void foo(int *arr[10]);  //the same as 7 (is it?)
10. void foo(int (*arr)[10]);//is the same as 8 (is it?)

11. void foo(int arr[]);    //is the same as 6 (is it?)
12. void foo(int arr[10]);  // is the same as 6 (is it?)

(I know this will need a lengthy explanation, sorry, I’m totally confused…)

Answer 1

The first important piece of information is that parameters whose type is a (bounded or unbounded) array of T are transformed to pointers to T. I.e. both int arr[] and int arr[10] are transformed to int * arr. Note that the transformation is only performed on top-level arrays, i.e. it doesn’t occur in int (*arr)[10], which is a pointer to an array of int.

Furthermore, things to the right of the identifier bind more closely than things to the left, i.e. int *arr[10] is an array, whereas int (*arr)[10] is a pointer.

Lastly, arrays of and pointers to references are invalid, as are pointers and references to unbounded arrays.

1. void foo(int &arr);        // can't pass, reference to int
2. void foo(int &arr[]);      // invalid, pointer to reference to int
3. void foo(int (&arr)[]);    // invalid, reference to unbounded array of int
4. void foo(int &arr[10]);    // invalid, pointer to reference to int
5. void foo(int (&arr)[10]);  // can pass, reference to an array of int

6. void foo(int *arr);        // can pass, pointer to int
7. void foo(int *arr[]);      // can't pass, pointer to pointer to int
8. void foo(int (*arr)[]);    // invalid, pointer to an unbounded array of int.
9. void foo(int *arr[10]);    // can't pass, pointer to pointer to int
10. void foo(int (*arr)[10]); // can't pass, pointer to array of int

11. void foo(int arr[]);      // can pass, pointer to int
12. void foo(int arr[10]);    // can pass, same as above

Using arr as an argument to foo will cause it to decay to a pointer to its first element — the value passed to foo will be of type int *. Note that you can pass &arr to number 10, in which case a value of type int (*)[10] would be passed and no decay would occur.