CODEWITHSUNDEEP
c++arrayspointersreferencepass-by-reference
caramel1995
caramel1995

Reputation: 3055

Reference to Pointer Type

#include <iostream>

using namespace std;
int main(void)
{
    int num[5];
    const int* &ref = num;

    return 0;
}

I've read a C++ book which mentioned if a reference variable is referencing to:

  1. a variable where the type is different but can be converted.
  2. a variable that is not a Lvalue.

As long as the referencing variable is declare as const, the above 2 cases will be solved by using a method where the compiler will create a storage and the value will be placed into it while the identifier of the referencing variable is treated as the identifier for that particular storage location . Below is the demonstration code .

Case 1 
#include <iostream>

using namespace std;
int main(void)
{
    int num = 5;
    const long long &ref = num; //the value 5 is place in a new storage where ref is the identifier

    return 0;
}
Case 2: 
#include <iostream>

using namespace std;
int main(void)
{
    int num = 5;
    const int &ref = num + 1; //expression num + 1 evaluated and store in new storage with identifier ref

    return 0;
}

Since this 2 cases is valid, how come the case inside The Code: is invalid?

My logic is since the name of the array when used will be converted to pointer to the first element of the array , thus the compiler should've spotted this is not a lvalue and a new storage will be created to store that address along and of course , the referencing variable name will be taken as the identifier for that location .

Note : I know this is slightly out of topic , but may I know whether an array name is Lvalue or not? Just a simple yes or no will do , since changing the code to int &ref = num I assume it's not a lvalue , but I just need further confirmation.

Thank you.

Upvotes: 2

Views: 744

Answers (1)

Steve Jessop
Steve Jessop

Reputation: 279315

You reference variable is not declared const.

There's a difference between const int * and int * const, and you've picked the wrong one.

Your example (ii) is invalid for the same reason, it should be const int &ref = num + 1;

For your Note, I'm not sure that a simple yes or no will do. A simple array name is an lvalue, referring to the array. However, in most contexts it decays to a pointer-to-first-element, which is an rvalue pointer.

Upvotes: 4

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