How to reference an array

[eker676]

I just bought "The C++ programming language" By Stroustrup and one of the exercises in it says to reference an array of 10 integers. Well here is the code I have but it doesn't work

     
    //reference to an array of 10 integers
  
    cout << endl << endl << "Values in nArray2 when referenced" << endl;
    
    int nArray2[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    
    int& pnArray2 = nArray2[10];
    
    for(;pnArray2 < 10; pnArray2++)
    {
                  cout << pnArray2 << endl;
    }

The code compiles but the count starts ouputing at -2 and it doesn't output the referenced array.

[firedraco]

I'm not very good with refrences, but is int& pnArray2 = nArray2[10]; correct? It seems like it would just get the refrence of the value at 10 and probably screw up the rest of your code.

[helios]

firedraco is right. References can't be incremented. Only the value the reference.
In any case, initializing it to nArray2[10] doesn't make much sense. You'd have to initialize it to nArray2, but then it'd be easier to just use the pointer directly.

EDIT: Oh, yeah. And there are better books out there for learning C++ than The C++ Programming Language.

[guestgulkan]

The OP got his array indexing mixed up.

However, I don't think you can make a reference of an array directly.


You can do something like this though:

        int nArray2[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    
    int* middleman = nArray2;

    int*& andy = middleman;
    

[eker676]

I didn't know you couldn't reference an entire array. I will try that reference to pointer method.
@helios Other books I've read teach the language in a bad order and are not indepth enough.

[jsmith]

If your only goal is to output the contents of the array, then I'd suggest forgoing references and just do it like this:

for( size_t i = 0; i < sizeof( nArray2 ); ++i )
   cout << nArray2[ i ] << endl;

[buffbill]

jssmith ...i don't think you can get the size of an array with 'sizeof'

[helios]

You can when the array is not dynamic.
Although you get the size in bytes. You should use sizeof(nArray2)/sizeof(int)

[jsmith]

Oops, yes, thanks for the fix helios.

Typically I use the template trick for this:

template< typename T, size_t N >
size_t ArraySize( T (& const)[ N ] )
{
    return N;
}

Edit: some older compilers (namely some versions of GCC) have issues with the parameter to this function and/or fail to match the const properly so sometimes you also have to declare the non-const version too:

template< typename T, size_t N >
size_t ArraySize( T (&)[ N ] )
{
    return N;
}

[helios]

I don't really see the point of doing that. The size is known at all times since it's constant. You might as well use that.

[jsmith]

The reason for using the template method is that it is hard(er) to get wrong. As opposed to the sizeof(array)/sizeof(array[0]) implementation which my original post clearly demonstrated is easily possible to get wrong :)

[helios]

I meant the point in getting the size at all, as opposed to just using the constant (10, in this case. Or DEF_BUFFER_SIZE for example).

[jsmith]

Yeah, this example is trivial enough that it's just easier to use the constant.

For a large app I'd use the template method as it avoids duplicating the constant (maintainability) and avoids having to use a #define.

[guestgulkan]

Well I think we have got away from the original posters question/problem.

An excercise in Stroustrup's book does the reader to declare various references, and one of them is a reference to an array.

Is it possible ( I think it may be trick question, and you can't declare such a reference).

[eker676]

Okay thank you for the help. I was just wondering if it could be done and how to do it. Maybe I Stroustrup worded the question wrong and he meant pointer. I know how to access arrays through pointers.

[jsmith]

To declare a reference to an array:

int array[ 10 ];
int (&array_ref)[ 10 ] = array;    // array_ref is now a reference to array 

The parentheses are required because otherwise the declaration

int &array_ref[ 10 ];

would attempt to declare an array of 10 references to ints.

[guestgulkan]

yes jsmith that works.
I thought I had tried every which way

[eker676]

I have saw the parentheses version of referencing and pointers but never paid much attention to it. Thank you all for clarifying everything.