WHat's up with VB6 and hex numbers ?

[cappy2112]

Why does VB6 do sign extension with hex numbers ???
Even if I use CLng(&hFFFF), VB displays -1

text3.text = "&HFFFF"
Label3.Caption = Str(CLng(Val(Text3.Text)))

FFFF is small enough where it should not cause sign extesnion, when referenced as a Long.

IS there another VB equivalent of typecasting ?

[Graff]

Have you tried referencing it as a double?

[cappy2112]

Originally posted by Graff
Have you tried referencing it as a double?

I think so. Is this what you mean -> CDBL() ??

All three of these text boxes contain &HFFFF

Label1.Caption = Str(Val(Text1.Text))
Label2.Caption = Str(CLng(Val(Text2.Text)))
Label3.Caption = Str(CDbl(Val(Text3.Text)))

And all three labels show -1

[Phenglai]

You cant include the &H into the text part of your text boxes, the compilier will think they are just text.

Place three labels and three text boxes on your form and past this code. this should work for you.

Private Sub Form_Load()
 
  Text1.Text = "FFFF"
  Text2.Text = "FFFFF"
  Text3.Text = "FFFFFF"
  
  Label1.Caption = CLng("&H" & Text1.Text)
  Label2.Caption = CLng("&H" & Text2.Text)
  Label3.Caption = CLng("&H" & Text3.Text)
 
End Sub

[BestCoder]

Originally posted by Phenglai
You cant include the &H into the text part of your text boxes, the compilier will think they are just text.

Place three labels and three text boxes on your form and past this code. this should work for you.

Sorry but I don't get the difference between:

Private Sub Form_Load()
 
  Text1.Text = "FFFF"
  Text2.Text = "FFFFF"
  Text3.Text = "FFFFFF"
  
  Label1.Caption = CLng("&H" & Text1.Text)
  Label2.Caption = CLng("&H" & Text2.Text)
  Label3.Caption = CLng("&H" & Text3.Text)
 
End Sub

And

Private Sub Form_Load()
 
  Text1.Text = "&HFFFF"
  Text2.Text = "&HFFFFF"
  Text3.Text = "&HFFFFFF"
  
  Label1.Caption = CLng(Text1.Text)
  Label2.Caption = CLng(Text2.Text)
  Label3.Caption = CLng(Text3.Text)
 
End Sub

[Phenglai]

Not much, personal preference really

[BestCoder]

Ok, your statement that "You cant include the &H into the text part of your text boxes, the compilier will think they are just text" (plus the fact that your code reflected that) confused me. A string is a string and whether it is concatenated or not will make no difference in converting hex to decimal.

[Phenglai]

I guess I wasn't clear on what I was trying to say. They had

CLng(&HFFFF)

and I was trying to say that for it to be a string for the compilier to recognize it

CLng("&HFFFF")

They needed to add the " "s in there so it would recognize it as a string.

Just didn't explain myself very well did I

[BestCoder]

BTW, cappy.

If I use this code:

Dim s1 As String
    Dim s2 As String
    Dim s3 As String
    
    s1 = "&HFFF"
    s2 = "&HFFFF"
    s3 = "&HFFFFF"
    
    Debug.Print CStr(CLng(s1))
    Debug.Print CStr(CLng(s2))
    Debug.Print CStr(CLng(s3))

it will yield the results:

4095
65535
1048575

Will that work for you?

[BestCoder]

OK, right, Phenglai. I gotcha now.

I hope either your method or mine should help cappy anyway.

[cappy2112]

Originally posted by BestCoder
BTW, cappy.

If I use this code:

Dim s1 As String
    Dim s2 As String
    Dim s3 As String
    
    s1 = "&HFFF"
    s2 = "&HFFFF"
    s3 = "&HFFFFF"
    
    Debug.Print CStr(CLng(s1))
    Debug.Print CStr(CLng(s2))
    Debug.Print CStr(CLng(s3))

it will yield the results:

4095
65535
1048575

Will that work for you?

I'm not sure yet- but at least it gives me another viewpoint.

I'm still trying to understand why &HFFFF gets converted to -1
in this situation

Dim doubnum As Double
Dim longnum As Long

doubnum = CDbl(&HFFFF)
longnum = CLng(&HFFFF)

Both values are = -1, when I break and evaluate the expression.

Is there a VB hex FAQ somewhere ? :-)))

[cappy2112]

[QUOTE]Originally posted by BestCoder
[B]BTW, cappy.

If I use this code:

Dim s1 As String
    Dim s2 As String
    Dim s3 As String
    
    s1 = "&HFFF"
    s2 = "&HFFFF"
    s3 = "&HFFFFF"
    
    Debug.Print CStr(CLng(s1))
    Debug.Print CStr(CLng(s2))
    Debug.Print CStr(CLng(s3))

it will yield the results:

4095
65535
1048575

dim d1 as double
dim L1 as long

d1= &HFFFF
L1= &HFFFF

printing these both yield -1.
WHY ??

[BestCoder]

If you haven't found your answer by tomorrow morning, I'll look at it again... my brain is getting fried right now.

[Graff]

DOH! Add a & after it haha I knew it was something simple.

Dim l1 as long

l1 = &HFFFF&

[BestCoder]

Oh right! I'm trying to remember how the signs work with hex, because &HFFF0 to &HFFFF is -16 to -1, just as &HFFFFFFF0 to &HFFFFFFFF is. I know it's something to with having negative numbers but I'm not sure how yet. I must look that up.

[cappy2112]

Originally posted by Graff
DOH! Add a & after it haha I knew it was something simple.

Dim l1 as long

l1 = &HFFFF&

THANKS !

But What does the 2nd & Do ?
Is this a documented feature ???

[cappy2112]

Originally posted by BestCoder
Oh right! I'm trying to remember how the signs work with hex, because &HFFF0 to &HFFFF is -16 to -1, just as &HFFFFFFF0 to &HFFFFFFFF is. I know it's something to with having negative numbers but I'm not sure how yet. I must look that up.

Wouldn't that depend on the data type bing integer, or long ?
Does VB just sign extend everything ???

[cappy2112]

Originally posted by Graff
DOH! Add a & after it haha I knew it was something simple.

Dim l1 as long

l1 = &HFFFF&

Ok- so now I know ho to prevent hex numbers from being converted to negative values, by using a trailing &, as in

Dim L1 as Long

L1=&HFF&
(evaluates as 255)


L1=&HFF
(evaluates as -1)


L1=&HFFFF&
(evaluates as 65535)

L1=&HFFFF
(evaluates as -1)


Since we are using a long data type in all of the assignmnets, why does VB represent the number as negative, when you don't use a trailing & ?

Is the trailing & just preventing sign extension ?

What is really happening behind the scenes ??

[beachbum]

Hi
dont worry about a trailing '&' as being the issue. It is all to do with how integers, longs etc are stored as negative or positive..

FFFF is 16 x 16 x 16 x 16 = 65536 which is the maximum 'capacity' of an integer. However, integers go from - 32768 to 32767. The first bit is used to show whether the number is positive or negative in that range... ie from 0 to 7 in the first bit gives u positive values and from 8 to F are for negative values.

&h0000 to &h7FFF = 0 to 32767
&h8000 to &hFFFF = -32768 to -1

The same concept is used with longs.

&h0000000 to &h7FFFFFFF = 0 to 2147483647
&h8000000 to & hFFFFFFF = -2147483648 to -1

[cappy2112]

Originally posted by beachbum
Hi
dont worry about a trailing '&' as being the issue. It is all to do with how integers, longs etc are stored as negative or positive..

FFFF is 16 x 16 x 16 x 16 = 65536 which is the maximum 'capacity' of an integer. However, integers go from - 32768 to 32767. The first bit is used to show whether the number is positive or negative in that range... ie from 0 to 7 in the first bit gives u positive values and from 8 to F are for negative values.

&h0000 to &h7FFF = 0 to 32767
&h8000 to &hFFFF = -32768 to -1

The same concept is used with longs.

&h0000000 to &h7FFFFFFF = 0 to 2147483647
&h8000000 to & hFFFFFFF = -2147483648 to -1

>>dont worry about a trailing '&' as being the issue.
You have no choice but to use them in VB, because Microsoft stuck us with with signed data types in VB.

If you don't use the trailing &, your logical tests will NOT come out correctly, among other problems.

[beachbum]

I dont see how ur logical tests can come out wrong if u know that both values are either signed or unsigned... it would only be an issue if u were logically comparing signed and unsigned values.

Ur original question was why &HFFFF became -1. Now u know

[cappy2112]

Originally posted by beachbum
I dont see how ur logical tests can come out wrong if u know that both values are either signed or unsigned... it would only be an issue if u were logically comparing signed and unsigned values.

Ur original question was why &HFFFF became -1. Now u know

Try this, it should explain it all. I've put one button, and one label on the form, but you dont even need them,. You could to this at form_load time to make it simpler.


Private Sub Command1_Click()
Dim L1 As Long
Dim L2 As Long


L1 = &HFFFF
L2 = &HFFFF&

If (L1 = L2) Then
Label1.Caption = "L1 = L2"
End If

If (L1 > L2) Then
Label1.Caption = "L1 > L2"
End If

If (L1 < L2) Then ' ONLY this test is true
Label1.Caption = "L1 < L2" ' this is what is displayed in label1
End If


End Sub

[Mad Compie]

What?

Dim L1 as Long 

L1=&HFF& 
'(evaluates as 255) 


L1=&HFF 
'(evaluates as -1) 


L1=&HFFFF& 
'(evaluates as 65535) 

L1=&HFFFF 
'(evaluates as -1)

I don't think that L1=&hFF equals -1??? Vb stores 2 bytes, so in this case &hff or &hff& always equals 255 and NOT -1!!!
Even if L1 was of type "Byte" it would still remain 255!

[beachbum]

Originally posted by cappy2112
Try this, it should explain it all. I've put one button, and one label on the form, but you dont even need them,. You could to this at form_load time to make it simpler.

Originally posted by beachbum
I dont see how ur logical tests can come out wrong if u know that both values are either signed or unsigned... it would only be an issue if u were logically comparing signed and unsigned values.

'Nuff said

[cappy2112]

>>I don't think that L1=&hFF equals -1??? Vb stores 2 bytes, so >>in this case &hff or &hff& always equals 255 and NOT -1!!!
>>Even if L1 was of type "Byte" it would still remain 255! [/B]

Dont think !
Type the code in, let VB show you how it is treating the numbers !

This is what matters, not what you think is happening !

[cappy2112]

Originally posted by beachbum
'Nuff said

Dont think !
Type the code in, let VB show you how it is treating the numbers !

This is what matters, not what you think is happening !

Just copy and paste the code I've posted above, it into a program,
before you prove you dont know what you're talking about, again.

[beachbum]

Oh gawd someone help me please!!! Your code is way too complicated for me to key in. I am still looking for the Any key and so will be a while before I find that Command1 thing. If you bother to read what I said instead of promoting your work of genius you will note that I said you would have no problems unless u were comparing signed values with unsigned values... and that is exactly what your vast production does. And you arent even using logical operators at all yet commenting as if you are. I dont need some wannabe fool telling me I dont know what I am talking about... there are plenty of others that will do that gladly

[cappy2112]

Throughout all of the this arguing, NOONE has explained what the trailing & does (as Microsoft would define it), or where it is documented in VB.

I can see what it is doing, and guess what's really happening under the covers, but I want to KNOW what it's called, and where it is documented (if at all).

[siphis83]

Thanks for the help guys! I can't believe my employer is still using VB 6....I should not have needed to find a forum from 2002...

[szlamany]

Throughout all of the this arguing, NOONE has explained what the trailing & does (as Microsoft would define it), or where it is documented in VB.

Google is your friend: vb6 hex number declaration

http://msdn.microsoft.com/en-us/library/s9cz43ek.aspx

[edit] this sentence is key:

Literals can also use the identifier type characters (%, &, @, !, #, $), as can variables, constants, and expressions. However, the literal type characters (S, I, L, D, F, R, C) can be used only with literals.

[/edit]

[Phill.W]

Since we are using a long data type in all of the assignments ...

Sure you are, but is there anything in your expression that tells VB that you're in any way, shape or form interested in Long values? No.

Faced with nothing to tell it otherwise, VB will pick the efficient option and stick to purely Integer arithmetic, hence your sign extension problem.

The Type Declaration Character, "&", tells VB that you explicitly want to work with a Long value at this point.

Regards, Phill W.

[Elroy]

This may already be answered as I didn't read everything, but &HFFFF is an integer (not a long). In fact, it's the integer = -1. You convert this integer to a long with CLng and you're just converting -1 as an integer to -1 as a long. If you want to truly have a hex LONG, then use the "&" declaration after the number. For instance, Debug.Print CLng(&HFFFF&) will print 65535. Or even Debug.Print &HFFFF& will print 65535, whereas Debug.Print &HFFFF will print -1. I hope this helps.

As a final thought, literals and also undeclared constants always go to the lowest possible type, unless specifically forced with one of the &!#%$@ type declarations.

[szlamany]

Google is your friend: vb6 hex number declaration

http://msdn.microsoft.com/en-us/library/s9cz43ek.aspx

[edit] this sentence is key: [/edit]

That's what the MSDN link I talks about.