reference doesn't work?

This is a discussion on reference doesn't work? within the C++ Programming forums, part of the General Programming Boards category; Hi folks, I have the following code of which I don't understand why the reference 'q' doesn't work: Code: #include ...

  1. #1
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    Unhappy reference doesn't work?

    Hi folks,

    I have the following code of which I don't understand why the reference 'q' doesn't work:

    Code:
    #include <cstdlib>
    #include <iostream>
    
    using namespace std;
    
    int main()
    {
        int n;
            
        cout<<"enter number of items: ";
        cin>>n;
        
        double *p = new double[n];
        double &q = *p;
            
        for (int i=0;i<n;i++) {
        q[i] = i;
        }
        
        cout<<"The first item is: "<<q[0]<<endl;
        cout<<"The last item is: "<<q[n-1]<<endl;
        
        
        delete [] p;
        
        
        system("PAUSE");
        return EXIT_SUCCESS;
    }
    The Bloodshed Dev C++ linker said for the line:
    Code:
     q[i] = i;
    Code:
    19 invalid types `double[int]' for array subscript
    However when I dropped the use of 'q', the program works fine. Is there anything wrong when I use a reference in such a context?

    Thanks a lot in advance...

  2. #2
    System Novice siavoshkc's Avatar
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    Don't do that.
    Last edited by siavoshkc; 08-09-2006 at 11:03 AM.
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  3. #3
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    I tried that, but...

    it only added the following error message:
    Code:
    14 invalid initialization of reference of type 'double&' from expression of type 'double*'

    Quote Originally Posted by siavoshkc
    double &q = *p;
    should be

    double &q = p;

  4. #4
    System Novice siavoshkc's Avatar
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    q is not array you can't use it as an array like q[index]. Use a pointer instead of a reference.

    double &q = *p; is ok
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  5. #5
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    Therefore...

    I have no means to reference a vector except using the pointer?

    Quote Originally Posted by siavoshkc
    q is not array you can't use it as an array like q[index]. Use a pointer instead of a reference.

  6. #6
    Cat without Hat CornedBee's Avatar
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    You could use a std::vector instead.
    All the buzzt!
    CornedBee

    "There is not now, nor has there ever been, nor will there ever be, any programming language in which it is the least bit difficult to write bad code."
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  7. #7
    System Novice siavoshkc's Avatar
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    Why not using pointer?
    Just change
    double &q = *p;
    to
    double *q=p;

    and everything will work fine.
    Last edited by siavoshkc; 08-09-2006 at 11:22 AM.
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  8. #8
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    actually

    it should be:
    Code:
     double *q=p;
    and everything works fine. Thanks!
    However I still don't get it: is reference operator not applicable in this context? (therefore it can only be used for referencing a single variable? That's uncomfortable...)


    Quote Originally Posted by siavoshkc
    Why not using pointer?
    Just change
    double &q = *p;
    to
    double *q=*p;

    and everything will work fine.

  9. #9
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    >> That's uncomfortable.
    Why? What do you expect to be able to do with a reference that you cannot do with a pointer?

  10. #10
    System Novice siavoshkc's Avatar
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    it should be:

    Code:
    double *q=p;
    Yes, sorry.
    However I still don't get it: is reference operator not applicable in this context? (therefore it can only be used for referencing a single variable? That's uncomfortable...)
    When you write
    double &q= *p;

    q will be exactly the same as *p. So q[n] (==*p[n]) has no meaning.

    [edit]
    People will correct me, but I think reference is used when you have a pointer but you want to see it as the type that the pointer is pointing to. It is usually used in function arguments when you want to pass by reference.
    Last edited by siavoshkc; 08-09-2006 at 11:21 AM.
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  11. #11
    (?<!re)tired Mario F.'s Avatar
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    If you really want a reference you should declare it like so:

    Code:
    double *&q = p;
    And it will work fine. q is a reference to a pointer to double.

    If you prefer a pointer you should declare it like so:

    Code:
    double *q = p
    And it will also work fine. q is a pointer to double.
    The programmer’s wife tells him: “Run to the store and pick up a loaf of bread. If they have eggs, get a dozen.”
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    Originally Posted by brewbuck:
    Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.

  12. #12
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    Hi

    I CAN do with a pointer; that's where the uneasiness comes from: why can't I do with a reference? What's wrong?

    Quote Originally Posted by Daved
    >> That's uncomfortable.
    Why? What do you expect to be able to do with a reference that you cannot do with a pointer?

  13. #13
    (?<!re)tired Mario F.'s Avatar
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    You can and should do it with a reference. The simple rule of thumb is "use references when you can, only use pointers when you must."
    The programmer’s wife tells him: “Run to the store and pick up a loaf of bread. If they have eggs, get a dozen.”
    The programmer comes home with 12 loaves of bread.


    Originally Posted by brewbuck:
    Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.

  14. #14
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    This

    ...looks opaque at a first glance though...

    [QUOTE=Mario F.]If you really want a reference you should declare it like so:

    Code:
    double *&q = p;
    And it will work fine. q is a reference to a pointer to double.

  15. #15
    System Novice siavoshkc's Avatar
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    Why you are insisting in using a reference?

    [edit]
    I am not sure about this but when we have a pointer named p;
    *p or p[0] is the value pointer is pointing to.
    p is the memory address that points to that value.
    &p is the address of pointer itself.
    But when you use a reference named q=*p.
    q is the value that p was pointing to.
    &q is the address of that value and is equal to p.
    Now we can't get the address of q itself. But there is an address.

    I mean
    Code:
    int n=2;
    int &r=n;
    Now r acts as same as n right? But it doesn't have the same type and it will take one pointer size location in memory more than n.
    Last edited by siavoshkc; 08-09-2006 at 11:43 AM.
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