Error: initialization of non-const reference type

This is a discussion on Error: initialization of non-const reference type within the C++ Programming forums, part of the General Programming Boards category; Hi! I have been thinking about this one the whole night but couldn't figure it out. I guess it was ...

  1. #1
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    Red face Error: initialization of non-const reference type

    Hi!

    I have been thinking about this one the whole night but couldn't figure it out. I guess it was my bad day (or night).

    So, I have written the following code (it does nothing useful, it is just for the illustration):

    Code:
    #include <iostream>
    using namespace std;
    
    class A {
      public:
      int* a;
      void change(int* & tmp) {a=tmp;}
      int* get() {return a;}
    };
    
    int main() {
      A temp;
      int* buff=new int(42);
      temp.change(buff);
      temp.change(temp.a);
      temp.change(temp.get());  //ERROR: Initialization of non-const reference type 'int *&', from rvalue of type 'int *', in passing argument 1 of 'A::change(int *&)'
      
      cout << temp.get() << endl;
      cout << temp.a << endl;
      
      return 0;
    }
    So I need to pass a pointer to a method by reference! It works fine in the line
    temp.change(temp.a);
    but when I use the get() method instead of just a:
    temp.change(temp.get());
    I get an error.

    Why???? I mean, both temp.a and temp.get() should return the same thing, shouldn't they???

    Now, I could have solved it by declaring the method A::change like this:
    void change(int* const & tmp) {a=tmp;}

    BUT I WANT TO BE ABLE TO CHANGE the variable tmp in this method. So the solution with const doesn't work for me.

    Any tips, explanations ?

    Thank you!

  2. #2
    and the hat of wrongness Salem's Avatar
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    > I mean, both temp.a and temp.get() should return the same thing
    They're the same thing by value (namely, an address), but a function return can only exist as an rvalue, and a reference is usually applied to an lvalue.

    So what's an lvalue and rvalue?
    left and right basically, as in left-hand side of an expression, or the right-hand side.

    A variable can exist as either, so you can say
    var = something; // var is lvalue
    something = var; // var is rvalue

    But for a function return, it must be an rvalue
    something = function(); // rvalue

    You can't for example say
    function() = something; // nope, it's not an lvalue

    That is basically what the error message is getting at - it's trying to create a reference to an lvalue, and all you have is an rvalue.

    My simplistic work-around is
    int *dummy = temp.get();
    temp.change(dummy);
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  3. #3
    Skunkmeister Stoned_Coder's Avatar
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    but a function return can only exist as an rvalue
    This is c++ not c. c++ has references. A function may return a reference and hence act as an lvalue.

    Now onto your problem.....

    int* get()

    This returns a pointer by value.....

    void change(int* & tmp)

    This takes a pointer by reference.

    temp.change(temp.get());

    This takes a pointer by reference but the pointer you give it is actually a temporary copy of the return value of temp.get(). The c++ standard says that you cant reference a temporary unless its const. Obviously this cant be const because you wish to change it so simple solution split the statement into 2 statements and use a named object instead of a temporary.
    Free the weed!! Class B to class C is not good enough!!
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