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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?
Code:int inumbers[10];
int *pinumbers = inumbers; //pointer to array
int &rinumbers= inumbers[0];// refrence to first element;
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:Code:int &rinumber=inumbers;// compiler error
Code:int* &rinumber=inumbers;// compiler error
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
Code:int inumbers[10];
int * &rinumbers=inumbers;
Use this instead:
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).Code:int * const & rnumbers = inumbers;
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.
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!
> 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.
>>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:
Code:int array[10];
int (&ra)[10] = array;
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 :)