Why '&' here means reference?-CodePudding

I'm reading C Primer and it really confused me when I read this:

int odd[] = {1, 3, 5, 7, 9};
int even[] = {0, 2, 4, 6, 8};
//return a pointer, which points an array with 5 int
decltype(odd) *arrPtr(int i)
{
    return (i % 2) ? &odd : &even;
}

int main()
{
    cout << "odd: " << odd << endl;
    cout << "even: " << even << endl;
    cout << arrPtr(3);
}

The result is:

odd: 0x472010
even: 0x472030
0x472010

What puzzles me is why there is a need for &. I mean, is it necessary to add a '&'?

I tried to delete & like this:

return (i % 2) ? odd : even;

And it just doesn't work:

[Error] cannot convert 'int*' to 'int (*)[5]' in return

Here's what I thought:

It can't be 'address of odd', for it will become a second-order pointer. So it must be a reference.(This is my opinion)

So the question is, why must it be a reference?

Can anyone explain this for me? Any answer is welcomed. Thanks a lot.

CodePudding user response：

decltype(odd) *arrPtr(int i)

This line is saying that you are going to return a pointer to an int array of size 5.

When you do

return (i % 2) ? odd : even;

You are trying to return an int array of size 5, not a pointer to it. Thus, the & in

return (i % 2) ? &odd : &even;

Is then taking the address of odd or even, and returning that as a pointer, which is what the function expects you to return.

Note: in C , you cannot return an array, but you can return pointers.

Also note: The type, when using & is int (*)[5], not int **, and the return type of the function is essentially int (*)[5], not int *

Also also note: As noted in the comments below, typically speaking, if you want to return just an integer array, you can use the return type of int *. An int[5] will be implicitly cast to a int * upon return. The use of the decltype(odd) is adding complexity.

CodePudding user response：

Why '&' here means reference?

It doesn't mean reference here. It is the addressof operator.

What puzzles me is why there is a need for &. I mean, is it necessary to add a '&'?

Because we are returning a pointer to the array. You get a pointer to the array using the addressof operator.

It can't be 'address of odd'

It can be, and it is.

for it will become a second-order pointer.

No. Pointer to array is not a second order pointer. Furthermore, second order pointers would be allowed, so this reasoning doesn't work.

So it must be a reference.

& means a reference only as a part of a type name (or function qualifier). It never means a reference as part of an expression.

I think I get part of it, but if I get the address of the pointer, it turns out 'int**'

You don't get the address of a pointer, so nothing "turns out int**".

You get the address of an array, so you get a pointer to the array.

how is it possible to assign it to 'int*'?

You don't assign to int*. You return the pointer to the array and insert it into an output stream.

CodePudding user response：

This decalres an array:

int odd[] = {1, 3, 5, 7, 9};

The type of the array is int [5]. Arrays can decay to a pointer to their first element, that is &odd[0]. And thats what is happening when you write: return (i % 2) ? odd : even;. The type of the pointer to first element is int*. The function, however, returns decltype(odd) *, a pointer to an array of 5 integers, that is int (*) [5]. And you can get a pointer to the array via the address-of operator: &odd.

Arrays are not pointers. Arrays can decay to pointers to their first element. This pointer is not the same as pointer to the whole array. &odd and &odd[0] point to the same memory location, but they are of different type. The first is a int (*) [5], pointer to array of 5 integers, the latter is int*, pointer to int.