InStr
returns an index. As a logical operator, And
wants to have Boolean
operands. Given Integer
operands, the And
operator is a bitwise operator - truth be told, these operators are always bitwise; we just dub them "logical" operators when the operands are Boolean
.
If InStr(ex_arr(i, ex_lc), "CriteriaA") Then
This condition is implicitly coercing the returned index into a Boolean
expression, leveraging the fact that any non-zero value will convert to True
.
Problems start when you bring logical/bitwise operators into the equation.
If InStr(ex_arr(i, ex_lc), "CriteriaA") And InStr(ex_arr(i, 4), "CriteriaB") Then dc(ex_arr(i, 2)) = ex_arr(i, 3)
Say the first InStr
returns 2
, and the second returns 1
. The If
expression becomes If 2 And 1 Then
, so 0
. That's zero, so the condition is false.
Wait, what?
Think of the binary representation of 2 vs that of 1:
2: 0010
1: 0001
AND: 0000
Bitwise-AND yields 0
, since none of the bits line up.
Stop abusing implicit type conversions, and be explicit about what you really mean. What you mean to be doing, is this:
If (InStr(ex_arr(i, ex_lc), "CriteriaA") > 0) And (InStr(ex_arr(i, 4), "CriteriaB") > 0) Then dc(ex_arr(i, 2)) = ex_arr(i, 3)
(redundant parentheses for illustrative purposes only)
Now this evaluates two Boolean
expressions, applies bitwise-AND to the two values, and correctly works as intended.
True: 1111
True: 1111
AND: 1111