Overloaded Operators, and Pointers

This is a discussion on Overloaded Operators, and Pointers within the C++ Programming forums, part of the General Programming Boards category; I have made a simple class which overloads an equals operator. So far, it does nothing important, becuase its just ...

  1. #1
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    Overloaded Operators, and Pointers

    I have made a simple class which overloads an equals operator. So far, it does nothing important, becuase its just a test.

    The class just looks like this:
    Code:
    class myoverloadAttempt
    {
    public:
    	int& val;
    
    	myoverloadAttempt(int& x) : val(x) {};
    
    	myoverloadAttempt& operator=(const int& x)
    	{ 
    		val=x;  
    		return *this;
    	};
    
    };
    and here is how I use it:
    Code:
    int myvar = 2;
    myoverloadAttempt o(myvar);
    o = 5;
        //myoverloadAttempt* o = new myoverloadAttempt(myvar);
        //o->operator =(5);
    std::cout<<myvar<<"\n";
    It works fine with those 2 lines commented out, ie using the class not as a pointer, but when I switch over and try to use it where o is a pointer to this class, o=5 no longer works. It says "error C2440: '=' : cannot convert from 'int' to 'myoverloadAttempt *'"
    When o is a pointer, It only worked when I used the line, o->operator=(5)
    .

    Is this always the case? Is there no way to use overloaded operators when dealing with a pointer?

  2. #2
    Captain Crash brewbuck's Avatar
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    Is this always the case? Is there no way to use overloaded operators when dealing with a pointer?
    Just write it this way instead:

    Code:
    *o = 5;
    Also, holding a reference to an outside object (like int& x) is generally a bad idea, since the object has no control over the external object's lifetime. I understand that this is just a little example, but still good to remember.

  3. #3
    Registered User Codeplug's Avatar
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    You need to comment out your other "o" when using your other "o"
    Code:
    int myvar = 2;
        //myoverloadAttempt o(myvar);
        //o = 5;
    myoverloadAttempt* o = new myoverloadAttempt(myvar);
    o->operator =(5);
    std::cout<<myvar<<"\n";
    gg

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    codeplug, I know, that was the only way it worked, but I wanted to just use o=5.

    And brewbuck, that works, so I will use that. Thanks, its not 100&#37; what I wanted, but what I wanted obviously doesn't exist, haha, so *o it is.

  5. #5
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by Codeplug View Post
    You need to comment out your other "o" when using your other "o"
    I think he was saying that that DOES work, but wanted to know how to avoid that ugly use of the "operator" keyword. The answer is to use '*' to dereference first.

  6. #6
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by esaptonor View Post
    And brewbuck, that works, so I will use that. Thanks, its not 100% what I wanted, but what I wanted obviously doesn't exist, haha, so *o it is.
    Nah, you can do what you want, I just don't see the point:

    Code:
    myObject *o = new myObject();
    myObject &oref = *o;
    oref = 5;
    Basically you turn the pointer into a reference, eliminating the need to use '*'. But really, I don't see the point. The star is not that big of a deal.

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    Ok, your right in that turning it into a reference is bad, I was hoping for some kind of modification to the operator overload part of the class...I assumed I was doing it wrong. But nah, I will stick with the * compared to having to turn it into a reference.

    Thanks a lot

  8. #8
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    Quote Originally Posted by esaptonor View Post
    And brewbuck, that works, so I will use that. Thanks, its not 100% what I wanted, but what I wanted obviously doesn't exist, haha, so *o it is.
    Doing just o = 5 tries to assign 5 to the pointer and not the object. A pointer needs to be dereferenced before it's accessed, so *o = 5 is the right way to go.
    You can also use const int& x, which is better since operator = doesn't change x.

  9. #9
    Registered User Codeplug's Avatar
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    >> o=5 no longer works

    Oh *that* part doesn't work

    You could do some crazy stuff *if* the language allowed for:
    Code:
    void operator=(myoverloadAttempt *p, int i) {*p = i;}
    Which it doesn't

    gg

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