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Home/ Questions/Q 4025542
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T10:51:34+00:00 2026-05-20T10:51:34+00:00

Consider the following code: class MyClass { template <typename Datatype> friend MyClass& operator<<(MyClass& MyClassReference,

  • 0

Consider the following code:

class MyClass
{
    template <typename Datatype>
    friend MyClass& operator<<(MyClass& MyClassReference, Datatype SomeData);
    // ...
};

template <typename Datatype>
MyClass& operator<<(MyClass& MyClassReference, Datatype SomeData)
{
    // ...
}

How can I define operator<< inside the class, rather than as a friend function? Something like this:

class MyClass
{
    // ...

    public:

    template <typename Datatype>
    MyCLass& operator<<(MyClass& MyClassReference, Datatype SomeData)
    {
        // ...
    }
};

The above code produces compilation errors because it accepts two arguments. Removing the MyClassReference argument fixes the errors, but I have code that relies on that argument. Is MyClassReference just the equivalent of *this?

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  1. Editorial Team
    Editorial Team
    2026-05-20T10:51:35+00:00Added an answer on May 20, 2026 at 10:51 am

    You have

    template <typename Datatype> MyClass& operator<<(MyClass& MyClassReference, Datatype SomeData);
    

    inside of the class. It is a method of the class MyClass. Non-static methods have an implicit parameter called the this pointer. The this pointer is a pointer to the object the method was called on. You do not need the MyClassReference parameter because the this pointer fulfills that purpose.

    Change that method declaration to

    template <typename Datatype> MyClass& operator<<(Datatype SomeData);
    

    .

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