subtracting base16

[cryoburned]

ok your idea to get the value of the string.. sort of works...

[code=vb]
Private Sub Form_Load()
hexfinal = "&H0000FFFF "
End Sub

Private Sub L2_Click()
If l2on = False Then
L2.ForeColor = &HFFFFFF
l2on = True
hexfinal = Hex(Val(hexfina l) - &HFFFF0100)
Text1.Text = "0000" & hexfinal
Text2.Text = "0010" & hexfinal
ElseIf l2on = True Then
L2.ForeColor = &H808080
l2on = False
hexfinal = Hex(Val(hexfina l) + Val("&H00000100 "))
Text1.Text = hexfinal
Text2.Text = hexfinal
End If
End Sub
[/code]

that was my last attemp..

the subtraction works.. i get FEFF like i wanted..

put the addition.. always results as the number 100 and not FFFF

[Killer42]

I'd like to try out your code, but I need you to fill in a couple of gaps. I assume Text1 and Text2 are textboxes, and hexfinal is obviously a string. But what are L2 and l2on?

[Killer42]

I think I should point out, though, that the result of VAL("FEFF") is zero, while VAL("&HFEFF") is -257 (or if you could get VB to treat it as unsigned, 65,279).

[cryoburned]

I'd like to try out your code, but I need you to fill in a couple of gaps. I assume Text1 and Text2 are textboxes, and hexfinal is obviously a string. But what are L2 and l2on?

L2 is a button, l2on is boolean

[cryoburned]

I think I should point out, though, that the result of VAL("FEFF") is zero, while VAL("&HFEFF") is -257 (or if you could get VB to treat it as unsigned, 65,279).

that did it.

Private Sub Form_Load()
hexfinal = "0000FFFF"
End Sub
Private Sub L2_Click()
If l2on = False Then
L2.ForeColor = &HFFFFFF
l2on = True
hexfinal = Hex(Val("&H" & hexfinal) - &HFFFF0100)
Text1.Text = "0000" & hexfinal
Text2.Text = "0010" & hexfinal
ElseIf l2on = True Then
L2.ForeColor = &H808080
l2on = False
hexfinal = Hex(Val(("&H" & hexfinal) + &H100))
Text1.Text = "0000" & hexfinal
Text2.Text = "0010" & hexfinal
End If
End Sub

works great now

[cryoburned]

hexfinal = Hex(Val(("&H" & hexfinal) + &H8000))

I believe you have too many parentheses. I think this should be...

Old version:
hexfinal = Hex(Val(("&H" & hexfinal) + &H8000))
New version:
hexfinal = Hex(Val("&H" & hexfinal) + &H8000)

[Killer42]

Sorry, my goof. I think I must have hit the Edit button instead of Reply. Damn! :(

[cryoburned]

Sorry, my goof. I think I must have hit the Edit button instead of Reply. Damn! :(

lo its cool.. but im still have the problem with 8000.. it goes to 7FFF in subtraction, but jumps to FFFFFFFF is addition.

[Killer42]

lo its cool.. but im still have the problem with 8000.. it goes to 7FFF in subtraction, but jumps to FFFFFFFF is addition.

Can you show us the exact code again? I think I zapped your last version when I did the accidental edit.

[QVeen72]

lo its cool.. but im still have the problem with 8000.. it goes to 7FFF in subtraction, but jumps to FFFFFFFF is addition.

Hi,

Late into this thread,
Actually &H8000 is equivalent to 32768
but if you use Val function or CDec Function, they Start Showing Negative Value.
Val("&H8000") = -32768
Val("&H8001") = -32767 And so on.
You can also use CLng, but again to a certain Limit ..

Better Idea will be to use CDbl Function..

CDbl("&H8000") = 32768
CDbl("&H8001") = 32769


Regards
Veena

[cryoburned]

Hi,

Late into this thread,
Actually &H8000 is equivalent to 32768
but if u use Val function or CDec Function, they Start Showing Negative Value.
Val("&H8000") = -32768
Val("&H8001") = -32767 And So On..
u can also use CLng, but again to a certain Limit ..

Better Idea will be to use CDbl Function..

CDbl("&H8000") = 32768
CDbl("&H8001") = 32769


Regards
Veena

that doesn't make a difference for this project.. but i got it.. i dont know why i had this problem but... now

[code=vb]
hexfinal = Hex(Val((("&H" & hexfinal) + &H8000) - &HFFFF0000))
[/code]

gets me FFFF like i wanted..

[Killer42]

hexfinal = Hex(Val((("&H" & hexfinal) + &H8000) - &HFFFF0000))

Hm... still seems as though you have too many parentheses there. But I guess if it works, it works. :)

Congratulations !

[Killer42]

... Better Idea will be to use CDbl Function ...

Thanks Veena, that's useful to know. I had no idea CDbl would produce a different result.