Thread: Point and Swap

  1. #1
    Registered User Kayoss's Avatar
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    Point and Swap

    This code successfully swaps the variables, but I'm wondering why the address for the first element (2) is greater than the address for the second element (3)?

    Code:
    #include <iostream>
    
    using namespace std;
    
    void swap( int * const, int * const ); //  prototype
    
    int main()
    {
    
       int array[2] = {2,3}; 
    
       int *arrayPTR = array; 
       int *arrayPTR1 = &array[1]; 
    
    
       cout << *arrayPTR << "  " << &arrayPTR << endl; // first element and address
       cout << *arrayPTR1 << "  " << &arrayPTR1 << endl; // second element and address
       cout << "SWAP" << endl;
    
    
       // the following code swaps elements in array, defined after   function main
    
       swap( &array[0], &array[0 + 1] );
    
    
       cout << *arrayPTR << "  " << &arrayPTR << endl;
       cout << *arrayPTR1 << "  " << &arrayPTR1 << endl;
    
    
    return 0;
    
    }
    
    // defines swap function, swaps values in array
    
    void swap( int * const element1PTR, int * const element2PTR )
    {
    	int hold = *element1PTR;
    	*element1PTR = *element2PTR;
    	*element2PTR = hold;
    
    } // end function swap
    THE redheaded stepchild.

  2. #2
    Registered User hk_mp5kpdw's Avatar
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    Who says data placed in memory has to start from the bottom and move up? It's entirely possible that it could be placed in the higher addresses first and move down from there.
    "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

  3. #3
    Registered User
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    On my system, when I run this program:
    Code:
    #include<iostream>
    using namespace std;
    
    int main()
    {
    	int arr[2]={2,3};
    	
    	cout<<arr<<endl
    		<<&arr[1]<<endl;
    
    	return 0;
    }
    I get this output:

    006BFDF0
    006BFDF4

  4. #4
    Registered User Kayoss's Avatar
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    Hmm I suppose it doesn't have to...it just has for me up until this point.

    I get:

    0012FF74
    0012FF70
    Last edited by Kayoss; 11-09-2005 at 03:45 PM.
    THE redheaded stepchild.

  5. #5
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    Who says data placed in memory has to start from the bottom and move up?
    Whaaa?? All the pictures in my book have the memory going from left to right?

  6. #6
    Just Lurking Dave_Sinkula's Avatar
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    Kayoss: Array members must be in an increasing order, but your program is not showing that -- it is showing the address of some pointers that hold the address you want.
    Code:
    #include <iostream>
    using namespace std;
    
    int main()
    {
       int array[2] = {2,3};
       cout << &array[0] << ' ' << array[0] << '\n';
       cout << &array[1] << ' ' << array[1] << '\n';
       return 0;
    }
    
    /* my output
    0012FF84 2
    0012FF88 3
    */
    [edit]
    Code:
    #include <iostream>
    using namespace std;
    
    void swap( int * const element1PTR, int * const element2PTR )
    {
    	int hold = *element1PTR;
    	*element1PTR = *element2PTR;
    	*element2PTR = hold;
    }
    
    int main()
    {
       int array[2]   = {2,3};
       int *arrayPTR  = array;
       int *arrayPTR1 = &array[1];
    
       cout << *arrayPTR  << "  " << arrayPTR << endl; // first element and address
       cout << *arrayPTR1 << "  " << arrayPTR1 << endl; // second element and address
       cout << "SWAP" << endl;
    
       // the following code swaps elements in array, defined after   function main
       swap( &array[0], &array[0 + 1] );
       cout << *arrayPTR  << "  " << arrayPTR << endl;
       cout << *arrayPTR1 << "  " << arrayPTR1 << endl;
    
       return 0;
    }
    
    /* my output
    2  0012FF84
    3  0012FF88
    SWAP
    3  0012FF84
    2  0012FF88
    */
    Last edited by Dave_Sinkula; 11-09-2005 at 04:05 PM.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  7. #7
    Registered User Kayoss's Avatar
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    Thank you Dave!
    THE redheaded stepchild.

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