How Do I Create New Instance Every New iteration

Question

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I Want To Create New Instance Every New iteration And I Am My Trying To concatenate Object Name Than It Throws Error

C#

static void Main(string[] args)
     {
         int i;
         char ch = 'n';
         while(ch=='n')
         {
             for(i=0;i<=10;i++)
             {   // I Want To Create New Instance Every New iteration And My Trying To concatenate  Object Name It Throws Error
                  lern "st"+i = new lern();
             }
         }

     }
 }
 class lern
 {
     public static int i = 0;
     public int m1,m2,m3;
     public string nm, sub;
    public lern ()
     {

     }
 }

Posted 4-Nov-15 20:39pm

Hitesh Jain

Updated 4-Nov-15 20:40pm

v2

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Comments

Gautham Prabhu K 5-Nov-15 4:53am

You cannot create variables at run time.

BillWoodruff 5-Nov-15 11:00am

Why do you have a loop that will never exit ?

5 solutions

Solution 1

No, Wrong way.
In C# do not allow to create variables names at runtime, this mean that your solution is impossible with C#.

You should have a look at documentation about arrays and resizeable lists.

Posted 4-Nov-15 20:59pm

Patrice T

Updated 4-Nov-15 21:01pm

v2

Comments

CPallini 5-Nov-15 3:11am

5.

Patrice T 5-Nov-15 3:24am

Thank You

Solution 2

You cannot create a variable like this

C#

lern "st"+i = new lern();

in c#. It will not even compile.

You have to do something like this:

C#

List<lern> lernList = new List<lern>();
for (i=0; i<=10; i++)
{
    lernList.Add(new lern());
}

That said, we have to take a look at your outer loop.

C#

while (ch=='n')
{
...
}

How will you get out of this loop?
Where do you change the value of ch?

Posted 4-Nov-15 20:59pm

George Jonsson

Updated 5-Nov-15 15:27pm

v3

Comments

Hitesh Jain 5-Nov-15 3:06am

while(ch=='n')
{
for(i=0;i<=10;i++)
{ // I Want To Create New Instance Every New iteration And My Trying To concatenate Object Name It Throws Error
lern "st"+i = new lern();

}
Console.WriteLine("Press s To Stop");
Console.ReadLine(ch);
}

Patrice T 5-Nov-15 3:25am

Use Improve question instead.

CPallini 5-Nov-15 3:12am

5.

George Jonsson 5-Nov-15 21:28pm

Thanks

Solution 3

You can't do that. You can't create variable names at run time, because the compiler can't check them! Even if you could, you couldn;t use them as they would go "out of scope" at the end of the list.

What you want to do is use an array, or a collection:

C#

private List<lern> lerns = new List<lern>();
static void Main(string[] args)
    {
    int i;
    char ch = 'n';
    while(ch=='n')
        {
        for(i=0;i<=10;i++)
            {   
            lern l = new lern();
            lerns.Add(l);
            }
        }
    }

Then in subsequent code you can refer to each instance separately:

C#

lern firstLern = lerns[0];
lern lastLern = lerns[lerns.Count - 1];

Posted 4-Nov-15 21:03pm

OriginalGriff

Comments

CPallini 5-Nov-15 3:12am

5.

Hitesh Jain 5-Nov-15 3:29am

Hey What Do You Mean By 5. ?

phil.o 5-Nov-15 3:47am

He means he gave the answer a note of 5, because he considers the answer to the question as perfectly valid.

Solution 5

You can not create a variable name at runtime, compile will not compile the code as

lern "st"+i = new lern();

The solution provided by pankaj is nic.

Posted 4-Nov-15 22:35pm

shashikantGuptaMCA

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PANKAJMAURYA · Accepted Answer · 2015-11-04T21:05:00

Try this

C#

static void Main(string[] args)
     {
List<lern> objList = new List<lern>();
       int i;
       char ch = 'a';
       while(ch<='n')
       {
           for(i=0;i<=10;i++)
           {
               objList.Add(new lern("st"+ch + i));
           }
           ch++;
       }
       string[] str = objList.Select(x => x.objID).ToArray();
       string result = string.Join("|", str);
}

class lern
    {
        public static int i = 0;
        public int m1, m2, m3;
        public string nm, sub;
        public string objID { get; set; }
        public lern(string strobjID)
        {
            this.objID = strobjID;
        }
    }</lern></lern>

How Do I Create New Instance Every New iteration

5 solutions

Solution 1

Solution 2

Solution 3

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Solution 4

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