[RESOLVED] What does a & after an argument/variable type signify?

I've run across this code and have seen similar code before, but I'm not sure what it means

Code:

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//--- file Person.h
. . .
class Person {
    private:
        char* _name;
        int  _id;
    public:
        Person& Person::operator=(const Person& p);
    . . .
}

//--- file Person.cpp
. . .
//=================================================== operator=
Person& Person::operator=(const Person& p) {
    if (this != &p) {  // make sure not same object
        delete [] _name;                    // Delete old name's memory.
        _name = new char[strlen(p._name)+1]; // Get new space
        strcpy(_name, p._name);              // Copy new name
        _id = p._id;                        // Copy id
    }
    return *this;    // Return ref for multiple assignment
}//end operator=

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Why is there a & after Person? What is it saying to the compiler?

Quote:

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Originally Posted by Brownhead

Why is there a & after Person? What is it saying to the compiler?

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That is the reference operator. The return value is a reference to a Person object, and the argument passed is a reference to a Person.

References, pointers, etc. should be discussed in any elementary C++ book.

Regards,

Paul McKenzie

Thank you :), I never learned about references.