Re: Why you should never use a const& parameter to initialize a const& member variable!

myFAQ 0815 - Why you should never use a const& parameter to initialize a
const& member variable!

Today, once again, I shot myself in the foot. I thought I'd share this.

Rule: You must never use a const-reference parameter to a constructor to
initialize a const-reference member-variable.
Reason: const& parameters bind to temporaries. You do not want to track
temporaries!
Solution: Use a const* parameter

If you want a const& member variable in a class to reference something,
then it has to be initialized in the ctor. But you must not use a const&
parameter to the ctor to initialize the member, because this parameter
would bind to a temporary and then you would be tracking the temporary
instead of the original value.

Example demonstrating the issue:
--------------------------------
#include <iostream>
using namespace std;

class Bad {
        int const& tracker_;

public:
        explicit Bad(int const& to_track)
        : tracker_(to_track)
        { }

        void print() {
                cout << "bad tracker_ is: " << tracker_ << endl;
        }

};

class Better {
        int const& tracker_;

public:
        explicit Better(int const* to_track)
        : tracker_(*to_track)
        { }

        void print() {
                cout << "better tracker_ is: " << tracker_ << endl;
        }

};

int f() {
        static int i = 1;
        i += 5;
        return i;

}

int main()
{
        int t = 100;
        char c = 32;

        Bad a1( f() ); // compiles: bad
        // Better b1( &(f()) ); - compiler error: good

        Bad a2( c ); // compiles: bad
        // Better b2( &c ); - compiler error: good

        Bad a3( t );
        Better b3( &t );

        t = 166;
        c = 64;
        t = f();

        a1.print(); // May crash or just print 6 (or whatever)
        // b1.print();
        a2.print(); // May crash or just print 32
        // b2.print();
        a3.print(); // OK
        b3.print(); // OK

        return 0;}

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