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;
}
--------------------------------

br,
Martin

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