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). :D

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??? :confused:

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!

> 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);

Quote:

---

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.