Getting Object reference not set to an instance of an object. why?

Question

1.00/5 (1 vote)

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C#

var split1 = str.Split(new[] { '~', ',', '!' });

 for (int i = 0; i < split1.Count(); i += 3)
            {
Label[] LabelName = new Label[split1.Count()];
                   LabelName[i].Name = split1[i].ToString();
}

C#

LabelName.Location = new Point(Convert.ToInt32(split1[i + 1]), Convert.ToInt32(split1[i + 2]));

Posted 8-Jan-16 17:19pm

Sathish km

Updated 8-Jan-16 18:13pm

v3

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2 solutions

Solution 1

It's because LabelName is an emtry array, you should initialize each element in the array before you can access its properties. Try this:

C#

var split1 = str.Split(new[] { '~', ',', '!' });
 
Label[] LabelName = (from sp in split1
                     select new Label{
                     Name = sp}).ToArray();

Posted 8-Jan-16 17:40pm

Member 11712753

Comments

Sathish km 9-Jan-16 0:17am

'string' does not contain a definition for 'Location' and no extension method 'Location' accepting a first argument of type 'string' could be found (are you missing a using directive or an assembly reference?

now getting error....

Sergey Alexandrovich Kryukov 9-Jan-16 3:19am

This code sample has a syntax bug, but "Location" is not mentioned anywhere. If it's a bug, it's your bug. :-)

Your LabelName is array, so it cannot have a member Location, only the array element has it. Problem solved.
But above, you report a different error not shown in your code sample. Now you try to find another non-existing member, String.Location. Do you want to have Location in all types? :-)

—SA

Sathish km 9-Jan-16 5:05am

yes

Sergey Alexandrovich Kryukov 9-Jan-16 5:44am

(Sign...)

Philippe Mori 10-Jan-16 19:55pm

Good start even though not complete as OP probably want to set location of each labels.

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Dave Kreskowiak · Accepted Answer · 2016-01-08T17:43:00

Solution 2

This is the easiest error to figure out. All you have to do is inspect variable contents when the code stops for an exception or hits a breakpoint you set.

The debugger is there to show you just how much you don't know about your own code and data.

Posted 8-Jan-16 17:43pm

Dave Kreskowiak

Comments

Sergey Alexandrovich Kryukov 9-Jan-16 3:17am

5ed. It's funny that, these days, a couple of members keep trolling some very correct answers on everyday basis. Perhaps the pattern is: too short or too general. How wrong! Yes, I know you don't care about it. You are perfectly right.
—SA