I’m looking at a 2001 function (vb6) and I’m having trouble understanding the If statement.
Sub x(a As byte) As byte
If (a AND &H80) then
TIA
I’m looking at a 2001 function (vb6) and I’m having trouble understanding the If statement.
Sub x(a As byte) As byte
If (a AND &H80) then
TIA
its a bitwise and
basically testing it the high bit is set
that code will NOT work in Xojo “as is”
basically it needs to be
If (a AND &H80) <> 0 then
That worked thanks.