CODEWITHSUNDEEP
c#.net
CoffeeCode
CoffeeCode

Reputation: 4314

basic c# question

why doesn't the element get swapped

public static void SwapArray(int[,] arr)
    {
        for (int i = 0; i < arr.GetLength(0); i++)
        {
            for (int j = 0; j < arr.GetLength(0); j++)
            {
                int temp = arr[i, j];
                arr[i, j] = arr[j, i];
                arr[j, i] = temp;
            }
        }
    }

even if the parameter is without a ref modifier the array doesn't change. a copy of the reference is passed as a parameter right?

Upvotes: 6

Views: 316

Answers (4)

n535
n535

Reputation: 5123

a copy of the reference is passed as a parameter right?

Arrays are passed by reference. 

SwapArray(ref int[,] arr)

Here you are passing a reference by reference (sorry, for tautology), this means, that you can even reassign a reference:

arr = new int [10,20];

Upvotes: 1

Rasshme
Rasshme

Reputation: 1641

try this.

I have changed the second for loop. u r actually swapping and again reswapping. so u stand where u were.

public static void SwapArray(int[,] arr) 
    { 
        for (int i = 0; i < arr.GetLength(0); i++) 
        { 
            for (int j = i+1; j < arr.GetLength(0); j++) 
            { 
                  int temp = arr[i, j]; 
                  arr[i, j] = arr[j, i]; 
                  arr[j, i] = temp; 
             } 
        } 
    }

Upvotes: 1

ReinierDG
ReinierDG

Reputation: 885

The second arr.GetLength(0) should be arr.GetLength(1). Because you want to use the 2nd dimension.

Upvotes: 4

Jens
Jens

Reputation: 25593

There is an error in your algorithm. For every i and j, your loop swaps arr[i,j] and arr[j,i] twice.

For example arr[3,1] gets swapped with arr[1,3] once for i=3, j=1 and once for i=1, j=3. So the result is the original matrix. You should change the j-loop to 

for (int j = 0; j < i; j++)

Upvotes: 21

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