C# error creating an array

Question

I need to assign values to a two-dimensional array. I can do it using multiple "myArray[x,y]" statements but I'd like to use another method (because I'll have arrays with many lines/columns) - see code:

int x;
x = 1;

string[,] myArray = new string[2, 2]; 


if (x == 1)
  {

    //does not work : why? Would be easier to populate a big array using this
    //myArray=
    //{
    // {"1", "1" },
    // {"1", "1" }
    //} ;

    //works, but I need above code to work if possible
    myArray[0, 0] = "1";
    myArray[0, 1] = "1";
    myArray[1, 0] = "1";
    myArray[1, 1] = "1";
    }

else if (x == 2)

    //does not work
    //myArray=
    //{
     //{"2", "2" },
     //{"2", "2" }
    //} ;

    myArray[0, 0] = "2";
    myArray[0, 1] = "2";
    myArray[1, 0] = "2";
    myArray[1, 1] = "2";
    }


MessageBox.Show(myArray[0,0]);

thanks

those type of array initializers only work when you're first declaring the variable. — DLeh
– DLeh, Commented Jan 6, 2015 at 21:55

musicfuel · Accepted Answer · 2015-01-06 21:59:02Z

2

I don't know if you are specifically looking to hardcode the values or not, but if you know the dimensions of the array to always be [2, 2] you can loop across all values of x you need.

var totalEntries = 10;
for (var x = 1; x <= totalEntries; x++) 
{
    for (var i = 0; i < 2; i++) 
    {
        for (var j = 0; j < 2; j++) 
        {
             myArray[i, j] = x.toString("G");
        }
    }
}

answered Jan 6, 2015 at 21:59

musicfuel

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Comments

Khalid Abdlqader · Accepted Answer · 2015-01-06 22:09:40Z

1

As you mentioned that you need to use it in that way, then you can do a workaround by declaring a temp variable initialize the values on it and then set the temp variable to the public one, like below:

int x;
x = 1;

string[,] myArray = new string[2, 2]; 

if (x == 1)
    {
         string[,] myArrayTemp = {     {"1", "1" },     {"1", "1" }    };
    }
else if (x == 2)
{
      string[,] myArrayTemp = {     {"2", "2" },     {"2", "2" }    };
      myArray = myArrayTemp;    
}

answered Jan 6, 2015 at 22:09

Khalid Abdlqader

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3 Comments

Mike Ness Over a year ago

I mostly agree with this answer except that it's not necessary to assign it to a temp variable. Using the new keyword in front of the set of braces will allow this syntax to work when the variable isn't being declared (i.e. myArray = new string[,] { {"1", "1" }, {"1", "1" } }).

Spacewalker Over a year ago

OK, I understood that I had a scope problem and that I've tried to declare my array with the same name twice in the same scope that is defined by the {...} pair. Thank you for the help.

Matt Over a year ago

Note: There is no need to create a temp variable, see here.

Tigran · Accepted Answer · 2015-01-06 21:54:28Z

1

Why don't just:

if(x == 1 || x == 2) {
  for(int row = 0; row < ROW_COUNT; row ++)
  {
     for(int col = 0; col < COL_COUNT; col++) 
     {
        myArray[row, col] = x.ToString();             
     }
  }
}

Not sure if if condition matters in your case.

If you are asking about something else, please clarify.

answered Jan 6, 2015 at 21:54

Tigran

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Comments

DLeh · Accepted Answer · 2015-01-06 21:57:09Z

1

You should also consider using loops to populate a large array.

var size = 1;
for(int i = 0; i <= size; i++)
{
    myArray[0, i] = x.ToString();
    myArray[i, 0] = x.ToString();
}

answered Jan 6, 2015 at 21:57

DLeh

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Comments

Matt · Accepted Answer · 2018-06-13 11:49:33Z

0

Referring to the other question you asked, simply try it this way:

    int x;
    x=1;

    string[,] myArray;
    switch (x)
    {
        case 1:
            myArray =  new string[,]{ { "1", "1" }, { "1", "1" } };     //OK
            break;

        case 2:
            myArray =  new string[,]{ { "2", "2" }, { "2", "2" } };     //OK
            break;
    }

You can't shorten these assignments, i.e. myArray = { { "2", "2" }, { "2", "2" } }; is not allowed (syntax error), because in C# you always need the new keyword to create a new object and a datatype which is string[,] (for two-dimensional arrays) if don't want to specify the array sizes beforehand (you don't need to count the elements this way).

edited Jun 13, 2018 at 11:49

answered Jan 9, 2015 at 15:10

Matt

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