Confusing pointer

This is a discussion on Confusing pointer within the C++ Programming forums, part of the General Programming Boards category; In my question paper there was a question to find output of the following program Code: #include<iostream.h> #include<conio.h> int main() ...

  1. #1
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    Confusing pointer

    In my question paper there was a question to find output of the following program

    Code:
    #include<iostream.h>
    #include<conio.h>
    
    int main()
    {
      int ia[3]={2,5,9};
      int *ptr=ia;
      
      for(int i=0;i<3;i++)
      {
        cout<<*(ia++);//Error Lvalue reqd.
        cout<<" "<<*ptr++;
      }
    getch();
    return(0);
    }
    On compiling it on my pc i am getting strange error l value reqd.Can anyone tell me why i am getting it. I also request you to compile it on your compiler as i am using old turbo c++.

  2. #2
    Devil's Advocate SlyMaelstrom's Avatar
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    If you're trying to say *ia + 1, then that isn't what *(ia++) means. Try changing it to *ia + 1

    When you say *(ia++) you're trying to use a non-lvalue as an lvalue.
    Last edited by SlyMaelstrom; 11-13-2005 at 02:25 AM.
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  3. #3
    Bond sunnypalsingh's Avatar
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    Array names are constant pointers...you can not alter them...read further from here

    http://www.fredosaurus.com/notes-cpp...spointers.html

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    thanks sunny,
    It did not strike me during paper.
    I have one more doubt.
    Code:
    int *p=new int[10];
    int p[i]=3;
    how can a pointer change to array here?

  5. #5
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    Code:
    int *p=new int[10];
    p[i]=3;
    This should work fine.

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    Quote Originally Posted by vaibhav
    thanks sunny,
    It did not strike me during paper.
    I have one more doubt.
    Code:
    int *p=new int[10];
    int p[i]=3;
    how can a pointer change to array here?
    Your declaring a new variable p in the second line, which should give you a compiler error saying something like "redefinition of p". A variable named p already exists in the first line.

  7. #7
    Bond sunnypalsingh's Avatar
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    Quote Originally Posted by vaibhav
    how can a pointer change to array here?
    Now you are dynamically allocating 10 bytes of memory which is pointed by pointer p.....and it is one of many ways to access the memory.

    p[i] is equivalent to *(p+i) and i[p]

    Note: Pointers are not arrays and arrays are not pointers

  8. #8
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    Assuming the the int size is 4 then allocating 10 ints allocates 40 bytes.

  9. #9
    Bond sunnypalsingh's Avatar
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    sorry for that.....and thanx for correction

  10. #10
    Frequently Quite Prolix dwks's Avatar
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    Quote Originally Posted by Quantum1024
    Code:
    int *p=new int[10];
    p[i]=3;
    This should work fine.
    You should probably also free the memory:
    Code:
    delete [] p;
    dwk

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