SSRS Iif Evaluates True and False - reporting-services

Question

MSDN states that SSRS will evaluate both the true and false part of an Iif statement regardless of which one is returned. This seems to make Iif completely useless in regard to error avoidance. In other words, you cannot use Iif to skirt around an error, so essentially any operation you choose to include must always be valid regardless of the conditions. Honestly, what good is this?
So my question is... Does SSRS have other ways of evaluating conditions besides Iif?
Here is my example. I just want to be able to return Left without grabbing the first character of the match.
=Iif
(
InStr(Fields!SearchField.Value, Fields!Criteria.Value) <= 1,
"",
Left(Fields!SearchField.Value, InStr(Fields!SearchField.Value, Fields!Criteria.Value)-1)
)
However, what is happening here is that InStr(Fields!Criteria.Value, Fields!Criteria.Value)-1 is evaluating to 0 in some cases, which is perfectly fine until the FALSE part of the statement tries to deduct 1 from it and pass it into the InStr function. InStr cannot accept -1 as the number of characters to return.
An oversimplification of this is as follows. Assume you have a situation where Value can never fall below 0 without throwing an error.
Iif (Value > 0, Value = Value -1, 0)
Trying to use Iif to force the value not to fall below 0 does not work because all of these statements get evaluated even if they do not meet the conditions.
Trying to use InStr to get an index on a match, and Left to build a substring based on that index fails because of this. I have no idea how to completely avoid the condition.

Answer 1

I too thought Switch would work, but upon testing it did not. As far as I can tell, custom code is the only way to go. I tested the function below which worked for my few test cases.
Public Function TruncateWord(ByVal str As String, ByVal criteria As String) As String
If str.Contains(criteria) Then
Return Left(str, InStr(str, criteria) - 1)
Else:
Return ""
End If
End Function
I tested with the below 5 basic strings, searching for "d", and got the following results:
+-----------------+
| String | Result |
+-----------------+
| asdf | as |
| asd | as |
| as | |
| da | |
| ad | a |
+-----------------+
So this appears to work for all 3 possible cases (InStr returns > 1, InStr returns 1, and InStr returns 0) from my limited testing.

Answer 2

Here is the final result from C Black's suggestion to use custom code. I had eventually hoped to use the segment of the string to format the match in a different color in the opposite column. I had to add some html tags. It works perfectly. Thank you all for your assistance.
Code block:
Public Function ParseMatch (ByVal FullString As String, ByVal Criteria As String) As String
Dim Segment(2) As String
Dim Result As String
If FullString.ToUpper.Contains(Criteria.ToUpper)
Segment(0) = Left(FullString, InStr(Ucase(FullString), Ucase(Criteria) )-1 )
Segment(1) = Criteria
Segment(2) = Right(FullString, Len(FullString) - Len(Segment(0)) - Len(Criteria))
Result = Segment(0) & "<b><FONT color=blue>" & Segment(1) & "</FONT></b>" & Segment(2)
Else
Result = FullString
End If
Return Result
End Function
Report cell:
=Code.ParseMatch(Fields!Defendants.Value, Fields!Firm_Client_Name.Value)
If the name is found in the list of defendants, it colors the text blue in that field and bolds it.

Answer 3

Use SWITCH
SWITCH stops evaluating expression as soon as the first True is found. Switch works with pairs (an expression to evaluate and a result if it's true). The final True acts like an else.
=SWITCH
(
InStr(Fields!SearchField.Value, Fields!Criteria.Value) <= 1, "",
True, Left(Fields!SearchField.Value, InStr(Fields!Criteria.Value, Fields!Criteria.Value)-1)
)

Answer 4

I rewrote it for readability:
=Switch
(
InStr(Fields!Defendants.Value, Fields!Firm_Client_Name.Value) = 0, "",
InStr(Fields!Defendants.Value, Fields!Firm_Client_Name.Value) = 1, "",
True, Left(Fields!Defendants.Value, InStr(Fields!Defendants.Value, Fields!Firm_Client_Name.Value)-1)
)
' 0 = Error
' 1 =
' >1 = substring based on criteria
1 and >1 are correct, but I still get the error when the InStr evaluates to 0.
The thing is, I have to tell the Left function -1 or it will return the first letter of the delimiter, which I do not want. Even though the condition of InStr(Fields!Defendants.Value, Fields!Firm_Client_Name.Value) = 0 is true, instead of returning "" for the column, it returns an error. This tells me it is still being evaluated despite it being outside of the specified condition.
If I omit the -1 within the Left function, no error results. Yet I get substring + first letter of delimiter.
I work with sensitive information, so I cannot give specific results of the strings.