Thread: Assigning value to a function with return type as ref!

  1. #1
    The Dragon Reborn
    Join Date
    Nov 2009
    Dublin, Ireland

    Assigning value to a function with return type as ref!

    #include <iostream>
    using namespace std ;
    class N
            int num ;
       int &geta(int n) 
          num = n ;   
            return num ;
        int get() { return num;} 
    int main()
        N n ;
        int &a = n.geta(12) ; 
        cout <<"First go: " << n.get() <<endl ; 
        a = 100;
        cout << "Second go: " <<  n.get() ; 
        n.geta(1)=19 ;
        cout << "Third go: " << n.get() ;
        return 0 ; 
     int &a = n.geta(12) ;
    i understand this piece of code. n.geta() returns a reference of member variable of object n to a
    so a <b>IS</b> num, an alias, but still num - and can be accessed like a normal variable which breaks the rule of encapsulation and access specifiers.
    as i have done with this code
    a = 100
    but I don't get this!!
    number 1 where i am not storing the reference of num in any reference variable.
    i was reading some stuff that n.geta() because it returns a reference can be used as a lvalue..
    but i don't get it! 
    where does the value that n.geta(1)  go to? 
     why is it that 1 which should be stored in num is overwritten by 19..or is it overwritten?
    Speaking of that
    how is it possible to initialize a value to a function as if it was a variable? 
    i keep trying to evaluate it and each time i fail: 
    geta(1) takes a parameter which is stored in num, so num is 1 
    it returns the value 1..but there is no reference variable in main to hold it..
    so it is floating somewhere in memory...and I just can't go on..
    what is 19? initialized to ?
    to num but how is that possible? 19 is not an argument, also .geta(1) is returning not storing...ack! can someone help!! THanks

    oh my God can someone tell me what is happening on this code?

  2. #2
    C++ Witch laserlight's Avatar
    Join Date
    Oct 2003
    Notice that n.geta(1) returns a reference. So, in a way, n.geta(1) is (another name for) n.num
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
    The Dragon Reborn
    Join Date
    Nov 2009
    Dublin, Ireland
    Thanks i get it :P

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