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program won't work

This is a discussion on program won't work within the C++ Programming forums, part of the General Programming Boards category; Why doesn't the following code work when we uncomment the function comp(double& t)? Isn't the funtion supposed to get called ...

  1. #1
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    program won't work

    Why doesn't the following code work when we uncomment the function comp(double& t)? Isn't the funtion supposed to get called when we do, comp c = n?

    Code:
    #include<iostream>
    #include<conio.h>
    #include<ctype.h>
    
    class comp
    {
    	double r,i;
    
    public:
    	comp (double rr = 0, double ii = 0)
    	{
    		 r = rr;
    		 i = ii;
    	}
    
    	/*comp(double &t)
    	{
    		r = t;
    		i = i;
    	}*/
    };
    
    
    int main()
    {
    	comp a(1.9,4.9), b(0.9,1.9);
    	double n = 0.1;
    	comp c = n;
    }

  2. #2
    Registered User manasij7479's Avatar
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    Quote Originally Posted by juice View Post
    Why doesn't the following code work when we uncomment the function comp(double& t)? Isn't the funtion supposed to get called when we do, comp c = n?
    Because what you're trying, needs an assignment operator.
    Like this:
    Code:
     void operator=(const double& t)
    {
         //Your code
    }
    Manasij Mukherjee | gcc-4.8.2 @Arch Linux
    Slow and Steady wins the race... if and only if :
    1.None of the other participants are fast and steady.
    2.The fast and unsteady suddenly falls asleep while running !



  3. #3
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    Quote Originally Posted by manasij7479 View Post
    Because what you're trying, needs an assignment operator.
    an assignment operator won't work either..

  4. #4
    Registered User manasij7479's Avatar
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    Quote Originally Posted by juice View Post
    an assignment operator won't work either..
    Really ? ...
    Then, perhaps you can enlighten me, why it is called an assignment operator !

    P.S: It works..
    Here is a small example:
    Code:
    class foo
    {
    public:
         int x;
         void operator=(int n){x=n;};
         foo(){}; 
    };
    Manasij Mukherjee | gcc-4.8.2 @Arch Linux
    Slow and Steady wins the race... if and only if :
    1.None of the other participants are fast and steady.
    2.The fast and unsteady suddenly falls asleep while running !



  5. #5
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    copy the code into your compiler, uncomment the function, introduce an overloaded assignment operator and check it out for yourself.

    You can use this assignment operator

    Code:
    void operator= (comp &t)
    {
    r = t;
    }

  6. #6
    Registered User manasij7479's Avatar
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    Quote Originally Posted by juice View Post
    copy the code into your compiler, uncomment the function, introduce an overloaded assignment operator and check it out for yourself.

    You can use this assignment operator

    Code:
    void operator= (comp &t)
    {
    r = t;
    }
    That is simply because the call becomes ambiguous with the constructor (as there are default values).
    That being the case, you don't even need an operator. The constructor will automatically be called for the assignment.(I don't know if that behaviour is standard, but my compiler seems to be satisfied.)
    Last edited by manasij7479; 02-15-2012 at 01:20 PM.
    Manasij Mukherjee | gcc-4.8.2 @Arch Linux
    Slow and Steady wins the race... if and only if :
    1.None of the other participants are fast and steady.
    2.The fast and unsteady suddenly falls asleep while running !



  7. #7
    Registered User hk_mp5kpdw's Avatar
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    Your default arguments are confusing things here. How is a single argument instantiation of the class going to resolve between the two constructors? Does it call the first constructor and use a default argument of 0 for the second, or does it call the single argument constructor? If you get rid of the default arguments, the code should compile.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

  8. #8
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    But when we say comp c = n, isn't a copy constructor supposed to get invoked..

  9. #9
    Registered User hk_mp5kpdw's Avatar
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    If n was another comp object then it would call the copy constructor. But, n is a double and that particular statement gets converted into a call to the constructor that expects a double argument (of which you had two available hence the confusion).

    Code:
    comp a;
    comp b = a;  // comp b(a)
    Here, since b and a are the same type, the copy constructor would be called.

    Code:
    double n = 9.0;
    comp a = n;  // comp a(n)
    Here, since n is a double, it tries to resolve which constructor to call based on ones that accept a double parameter. You have two of them, the one with the default arguments and the one with the one double argument. Either are possible options and the compiler complains.
    rags_to_riches likes this.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

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