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Unit 5: VB String and Operators

’ SearchFromRight is True, the returned string is that string to the left of the
’ LAST occurrance of TrimToChar. If TrimToChar is not found in the string,
’ the entire Text string is returned. TrimChar may be more than one character.
’ Comparison is done in Text mode (case does not matter).
’ If TrimChar is an empty string, the entire Text string is returned.
’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’’
Dim Pos As Integer
‘ Test to see if TrimChar is vbNullString. If so, return the whole string. If we
‘ don’t test here and used InStr as in the next logic block, an empty string would
‘ be returned.
If TrimChar = vbNullString Then
TrimToChar = Text
Exit Function
End If
‘ find the position in Text of TrimChar
If SearchFromRight = True Then
‘ search right-to-left
Pos = InStrRev(Text, TrimChar, -1, vbTextCompare)
Else
’ search left-to-right
Pos = InStr(1, Text, TrimChar, vbTextCompare)
End If
’ return the sub string
If Pos > 0 Then
TrimToChar = Left(Text, Pos - 1)
Else
TrimToChar = Text
End If
End Function

Perhaps the biggest bottleneck is that VB makes copies of the string data when
doing some of the operations. Even when you’re just reading strings (and not
planning to make any modifications), you can easily end up making a large
number of copies. The copying costs you time if string processing is an
intensive part of your program. Another reason is that some of the widely
used functions are implemented in a non-straightforward way. They may be
doing more work than what is required for your task. Fortunately, you can
often replace an advanced functions with a simpler and faster alternative.

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