Closed paul35621 closed 1 year ago
It seems that you can remove | (isWin8 & 2) * 4) in KeyDecoder.cs. Since isWin8 = ... & 1 it's always true that isWin8 & 2 == 0. Right?
| (isWin8 & 2) * 4)
isWin8 = ... & 1
isWin8 & 2 == 0
Exactly! isWin8 is either 0 or 1, hence isWin8 & 2 is always 0. But in that case, what is the variable even doing there?
isWin8
0
1
isWin8 & 2
It is just about the part of a code, not functionality. No need to fix.
It seems that you can remove
| (isWin8 & 2) * 4)
in KeyDecoder.cs. SinceisWin8 = ... & 1
it's always true thatisWin8 & 2 == 0
. Right?