how to convert a string value to int in C#

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How do OI convert string value to integer in C#?

I need to convert IDs here to integer.

string[] IDs = hdnFldSelectedValues.Value.Trim().Split('|');

Posted 2-Feb-11 18:34pm

RKT S

Updated 2-Feb-11 19:13pm

Chris Maunder772.8K

v3

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Sergey Alexandrovich Kryukov · Accepted Answer · 2011-02-02T18:48:00

This is how:

C#

string[] IDs = hdnFldSelectedValues.Trim().Split('|');
int[] Values = new int[IDs.Length];
for(int index = 0; index < Values.Length; index++)
    Values[index] = int.Parse(IDs[index]);

This code will throw exception on first invalid numeric format of out of range. If invalid format should be converted to a special value indicating error (not recommended) use int.TryParse:

C#

string[] IDs = hdnFldSelectedValues.Trim().Split('|');
int[] Values = new int[IDs.Length];
for(int index = 0; index < Values.Length; index++) {
    int value;
    if (!int.TryParse(IDs[index], out value))
        value = -1;
    Values[index] = value;
} //loop index

In last example, failure to parse a string will be converted to -1.
Again, throwing exception is better.

—SA

Igor Kushnarev · Accepted Answer · 2011-02-02T18:51:00

Solution 2

C#

List<int> list = new List<int>();
foreach (string id in IDs)
{
    int result;
    if (int.TryParse(id, out result))
        list.Add(result);
}

Posted 2-Feb-11 18:51pm

Igor Kushnarev

Comments

Sergey Alexandrovich Kryukov 3-Feb-11 1:09am

There is a bug: what happens if TryParse returns false? The variable result is not assigned.
Unfortunately, List is a huge overkill for such a simple task, the array is more then enough. TryParse suggests ill practice, should be advised better variant with exception. OP would not get explanation what happens with invalid numeric format. Sorry, pretty bad answer. I normally don't accept such quality of code.
Your arguments are welcome.
--SA

Chris Maunder 3-Feb-11 1:15am

An out paramter will always be assigned a value. Even so, I follow the pattern of

if (!int.TryParse(..., out variable))
variable = 0;

Sergey Alexandrovich Kryukov 3-Feb-11 1:28am

You're right. In this code there is not control over the int value used when TryParse fails.
"Always" means: in the case of failure variable = default(int), which is not always what can be indended.
--SA

Sergey Alexandrovich Kryukov 3-Feb-11 1:34am

Thank you Chris, I fixed a bug.
--SA