string startIndex 不能大于字符串的长度。参数名称：vb.net 中的 startIndex
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startIndex cannot be larger than length of string. Parameter name: startIndex in vb.net
提问by user2784251
I have a piece of code that generates random characters. The problem is, every once in a while, it returns an error:
"startIndex cannot be larger than length of string.
Parameter name: startIndex"
我有一段生成随机字符的代码。问题是，每隔一段时间，它就会返回一个错误：
“startIndex 不能大于字符串的长度。参数名称：startIndex”
How do I prevent this kind of error from happening?
我如何防止这种错误的发生？
Here's my code:
这是我的代码：
Friend Function gentCtrlChar()
Dim ran As New Random
Dim alpha As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
Dim alpha2 As String = "ZYXWVUTSRQPONMLKJIHGFEDCBA"
Dim rdm As New Random
Dim genChar As String = ""
For i As Integer = 1 To 52
Dim selChar As Integer = rdm.Next(1, 28)
Dim selChar2 As Integer = rdm.Next(1, 28)
genChar = genChar + "" + alpha.Substring(selChar, 1) + "" + alpha2.Substring(selChar2, 1)
On Error Resume Next
Exit For
Next
Return genChar
End Function
as you can see, I tried putting the "On Error Resume Next" hoping that somehow, this will take care of the error for me. But sadly, It doesn't do it's job. Or am I using it the wrong way or for the wrong situation?
如您所见，我尝试将“On Error Resume Next”放在后面，希望以某种方式解决我的错误。但遗憾的是，它并没有完成它的工作。还是我以错误的方式或在错误的情况下使用它？
Any help?
有什么帮助吗？
Thanks!
谢谢！
回答by BWS
this code:
这段代码：
Dim selChar As Integer = rdm.Next(1, 28)
will sometimes return a number that is longer (27 or 28) than the length of this string:
有时会返回一个比这个字符串的长度更长的数字（27 或 28）：
Dim alpha As String = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" (only 26 characters long)
so, this is invalid when selChar is 26 or more.
因此，当 selChar 为 26 或更多时，这是无效的。
alpha.Substring(selChar, 1)
Easiest fix is:
最简单的修复方法是：
Dim selChar As Integer = rdm.Next(0, alpha.Length)
Dim selChar2 As Integer = rdm.Next(0, alpha2.Length)
回答by jcwrequests
Try this way. I think its cleaner and easy to understand. A - Z is the same as 65 - 90 on the ascii map so its very easy to convert an integer into a Char value. Then we just use the string builder to make this easier to read.
试试这个方法。我认为它更清晰易懂。A - Z 与 ascii 映射上的 65 - 90 相同，因此很容易将整数转换为 Char 值。然后我们只使用字符串构建器来使它更容易阅读。
Dim rdm As New Random
Dim genChar As New StringBuilder()
For i As Integer = 1 To 52
Dim selChar As Char = Chr(rdm.Next(65, 90))
Dim selChar2 As Char = Chr(rdm.Next(65, 90))
genChar.Append(selChar)
genChar.Append(selChar2)
Next
Return genChar.ToString
