Technicalities involved with array of references

Question

I wrote this code for the following problem. My compiler is givin me the following error- [Error] declaration of 'a' as array of references. What is the problem?

void repfunc(int& a[],int size){

    for(int i=size-1;i>=0;--i){

            if(a[i]==0){
                a[i]=1;
                return;
            }

        for(int j=i-1;j>=0;--ji){

            if(a[i]==a[j]&&a[i]!=9){
                ++a[i];
                a[i]%=10;
                return;
            }
            else if(a[i]==a[j]&&a[i]==9){
                a[j-1]++;
                a[j-1]%=10;

                    FOR(k, 1,size-j){
                        a[k]=k;
                    }

                return;
            }
        }

    }

}

You can't have an array of references, that's the problem (references are not objects). — jrok, Nov 27 '13 at 13:49
[Why arrays of references are illegal?](http://stackoverflow.com/q/1164266/96780) — Daniel Daranas, Nov 27 '13 at 13:49
To find out why, Please follow the [link](http://stackoverflow.com/q/1164266/96780) in my comment above. Scroll down to the answers. About "what do I do", well, you still have two arrays to choose - array of objects and array of pointers. — Daniel Daranas, Nov 27 '13 at 13:51
then how do i pass an array to a function which can edit the array, without declaring the array global? — Soumadeep Saha, Nov 27 '13 at 13:52

score 6 · Accepted Answer · answered Nov 27 '13 at 13:53

6

As the error says, you can't have an array of references. It looks like you're actually trying to pass an array by reference. You can (sort of) do that by changing it to take a pointer to the start of the array:

void repfunc(int a[], int size)  // equivalent to "int * a"

Since arrays are convertible to pointers, this will do the right thing if the argument is an array

int a[4] = {1,2,3,4};
repfunc(a, sizeof a / sizeof a[0]);

Alternatively, you could accept a reference to an array of the correct size:

template <size_t size>
repfunc(int (&a)[size])

allowing the size to be deduced automatically

int a[4] = {1,2,3,4};
repfunc(a);

but note that this will only work if a is actually an array; not, for example an pointer to a dynamically allocated array.

answered Nov 27 '13 at 13:53

Mike Seymour

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@Paranaix: Indeed, I've added a note about that. – Mike Seymour Nov 27 '13 at 13:55
Careful, `sizeof a` is misleading there. – R. Martinho Fernandes Nov 27 '13 at 13:56
@R.MartinhoFernandes: Good point, I was typing faster than I was thinking. – Mike Seymour Nov 27 '13 at 13:56
if I do int a; std::cin>>a; int array[a]; is it dynamic? and if i pass an array by pointers will the changes be made on the array or on a copy of the array? – Soumadeep Saha Nov 27 '13 at 13:57
1

@SoumadeepSaha: No, that's an automatic array; it's dynamic if you use `new`, in which case you only have a pointer, not the array itself. If you pass a pointer, the changes will be made to the array, since that's what the pointer points to. – Mike Seymour Nov 27 '13 at 14:00
1

@SoumadeepSaha `int array[a]` won't compile if `a` is not a constant. – Johan Nov 27 '13 at 14:00
@Johan i use that all the time. Speaking from experience I think it will. – Soumadeep Saha Nov 27 '13 at 14:03
1

@SoumadeepSaha: Some compilers accept variable-length arrays as a non-standard extension. Don't use them if you want to write portable code. – Mike Seymour Nov 27 '13 at 14:04
@Mike Seymore i will keep that in mind. I think a vector here would have been more neat. But i am a newbie, any links to learning about them is welcome. – Soumadeep Saha Nov 27 '13 at 14:08
@SoumadeepSaha I did not even know it was accepted by some compilers.:) – Johan Nov 27 '13 at 14:16

score 1 · Answer 2 · edited May 23 '17 at 11:52

C++ Standard 8.3.2/4 clearly says that:

There shall be no references to references, no arrays of references, and no pointers to references

Actually a reference must always refer to something - there is no such thing as a null reference. This is why there can be no arrays of references, as there is no way to default instantiate references inside an array to a meaningful value. -Source is Here.

And also By origin array is accessed via pointer arithmatic. So if we want to support array of reference we have to support pointer to reference, as per standard which is not possible. These are reason array of reference is not possible. Which is what you were doing.

And as Mike Seymour mentioned : void repfunc(int a[], int size) is perfectly ok for your case.

Technicalities involved with array of references

2 Answers2