Re: A const behavior I don't understand

From:
Triple-DES <DenPlettfrie@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 19 Jun 2008 23:44:03 -0700 (PDT)
Message-ID:
<b8cf33ee-7a5e-4a53-87fa-d1cd283dd3a4@y21g2000hsf.googlegroups.com>
On 19 Jun, 17:54, Ben Thomas <ben.tho...@wickedstudios.com> wrote:

Hello,

I have the following code which I don't understand why it works :

#include <iostream>
using namespace std;

void DontWork (unsigned int& i) {
        cout << i << endl;

}

void Work (unsigned int const& i) {
        cout << i << endl;

}

void main () {
        signed int i = 1;
        // DontWork (i);
        Work (i);

}

What I don't understand is why the compiler can find an implicit
conversion when I add the const modifier, but it is unable to do so
without it.


In the process of converting from 'int' to 'unsigned int' a temporary
object is created. You cannot bind a non-const reference to a
temporary object.

Consider the example:
#include <iostream>
int main ()
{
  int i = 1;
  // fails: binding a non-const ref to a temporary
  unsigned& uref = i;

  // ok, binding const ref to temporary
  const unsigned& curef = i;

  i = 2;
  if( i != curef)
    std::cout << "curef is not bound to i";
}

You may think that the reference curef is bound to i, but it is not,
it is bound to an unnamed temporary object, as illustrated by the
example.

DP

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