refrence to an array of int?

[curlious]

Bjarne Stroustrup asks in C++ Programming Language 3rd edition exercise 5.9 1. to declare a refrence to an array of 10 integers. Is this possible. I know you can declare pointer to an array but a refrence?

int inumbers[10];
int *pinumbers = inumbers;  //pointer to array

int &rinumbers= inumbers[0];// refrence to first element;

int &rinumber=inumbers;// compiler error

[Zach L.]

int &rinumber=inumbers;// compiler error

The reason for the error is that you are trying to assign to a reference to an int, not an int* (i.e. it is roughly equivalent to 'char& r = inumbers', the types are incompatible). To create the reference, you need:

int* &rinumber=inumbers;// compiler error

[curlious]

ok thanks thought I was going bonkers before that explanation was under the assumption from the compiler error that you could not initialize a refrence with a pointer.

I still get the compiler error:
declare.cc:17: error: invalid initialization of non-const reference of type '
int*&' from a temporary of type 'int*'

with

int inumbers[10];
int * &rinumbers=inumbers;

[thefroggy]

Use this instead:

int * const & rnumbers = inumbers;

The reason you need a const is that an array cannot be modified. It would be illegal to take the above reference and increment it (as this would modify the location of the array).

By the way, which compiler are you using? I don't agree with the error message that you're receiving. It seems like it's reached a default case in trying to diagnose a failure to initialize a reference.

[curlious]

I am using g++ under linux, I am not sure what version, but am fairly sure it is up to date as I am using fedora.

I didn't realize you could increment a refrence anyhow like pointer arithmatic.

I assume I could use the refrence in the form
rinumbers[index]=10;
correct?

I am going to go implement your solution and see what the compiler tells me. Hopefully nothing!

[thefroggy]

> I didn't realize you could increment a refrence anyhow like pointer arithmatic.

You can if you're referencing a pointer-- but not an array. But remember that a reference is just an alias. If you reference a pointer and then increment it, the pointer itself will also change. This is the reason why a reference to an array must be a reference to const: you are not allowed to change the location of an array.


> I assume I could use the refrence in the form

Yes, that will work as you would expect.

[Prelude]

>>to declare a refrence to an array of 10 integers
>int * const & rnumbers = inumbers;<and similar answers>
You realize that every solution posted already fails to meet the requirement? All of these are references to pointers, not references to arrays. To get a true reference to an array, you would do this:

int array[10];
int (&ra)[10] = array;

[curlious]

Interesting Prelude, looking at the solutions closely I see the difference. Even what is supposed to be basic declarations can be difficult. My first attempt at the declaration was way off. In my attempt I forgot that an array name is both a pointer (though I intuitivly new this when using another pointer) and that it is constant. What threw me is my use of a pointer r, which I assumed would be similar with refrence, but like you explained a refrence is just an alias and a refrence to a constant pointer in the case of an array.

We all missed the fact that now we had a refrence to a constant pointer, and not actually to an array. The only difference I could see in the result between your solution and previous solutions is in the use of pointer aritmatic which, I am guessing is forbidden with your solution. I will go try it and check. No I was wrong of course. A refrence to a const pointer cant be changed nor can preludes soltution. doh.

I think I have a better grip on both pointers, arrays and refrences because of everyones help and this little exercise thanks. (Of course I thought I knew it all before trying this