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Home/ Questions/Q 601869
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T16:45:42+00:00 2026-05-13T16:45:42+00:00

I was told the reference variable must be initialized in the initialization list, but

  • 0

I was told the reference variable must be initialized in the initialization list, but why this is wrong?

   class Foo
    {
    public: 
        Foo():x(0) {      
         y = 1;
        }
    private:
        int& x;
        int y;
    };

Because 0 is a temporary object? If so, what kind of object can reference be bound? The object which can take an address?

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  1. Editorial Team
    Editorial Team
    2026-05-13T16:45:42+00:00Added an answer on May 13, 2026 at 4:45 pm

    0 is not an lvalue, it’s an rvalue. You cannot modify it, but you’re trying to bind to a reference where it could be modified.

    If you make your reference const, it will work as expected. Consider this:

    int& x = 0;
    x = 1; // wtf :(
    

    This obviously is a no-go. But const&‘s can be bound to temporaries (rvalues):

    const int& x = 0;
    x = 1; // protected :) [won't compile]
    

    Note that the life-time of the temporary is ended at the completion of the constructor. If you make static-storage for your constant, you’ll be safe:

    class Foo
    {
    public:
        static const int Zero = 0;
    
        Foo() : x(Zero) // Zero has storage
        {
            y = 1;
        }
    private:
        const int& x;
        int y;
    };
    
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