Look at the code below. All three variables are initialized to the same
value, -9223372036854775808. The first two work OK but the third throws a
overflow error. Why? (Framework v1.0.3705)

'-- OK, no overflow
Dim i as Long = Long.MinValue
Dim k as Long = CLng(-2 ^ 63)

'-- Overflow error
Dim m as Long = -9223372036854775808

"Brian" <(E-Mail Removed)> escreveu na mensagem
news:(E-Mail Removed)...
> Look at the code below. All three variables are initialized to the same
> value, -9223372036854775808. The first two work OK but the third throws a
> overflow error. Why? (Framework v1.0.3705)
>
> '-- OK, no overflow
> Dim i as Long = Long.MinValue
> Dim k as Long = CLng(-2 ^ 63)
>
> '-- Overflow error
> Dim m as Long = -9223372036854775808
>
>

You should add the literal type charecter 'L'.
Something like -9223372036854775808L


----
Rafael Pivato

Nope, already tried to append the L.


"Rafael Pivato" <(E-Mail Removed)> wrote in message
news:%(E-Mail Removed)...
>
> "Brian" <(E-Mail Removed)> escreveu na mensagem
> news:(E-Mail Removed)...
> > Look at the code below. All three variables are initialized to the same
> > value, -9223372036854775808. The first two work OK but the third throws
a
> > overflow error. Why? (Framework v1.0.3705)
> >
> > '-- OK, no overflow
> > Dim i as Long = Long.MinValue
> > Dim k as Long = CLng(-2 ^ 63)
> >
> > '-- Overflow error
> > Dim m as Long = -9223372036854775808
> >
> >
>
> You should add the literal type charecter 'L'.
> Something like -9223372036854775808L
>
>
> ----
> Rafael Pivato
>
>

The 1st example returns a Long that is well in range.
The 2nd example tries to convert the number to long since you explicitly
told it to do so using CLng(operation). The computer doesn't try figure
out exact value at compile time and because you told it to use CLng() to
change whatever number resulted into a long it assumes you know what you
are doing so it doesn't complain.
The 3nd example is obvious to the compiler the number you typed in will
never behave as the value as it is entered so it's an error.

"Rafael Pivato" <(E-Mail Removed)> schrieb
>
> "Brian" <(E-Mail Removed)> escreveu na mensagem
> news:(E-Mail Removed)...
> > Look at the code below. All three variables are initialized to the
> > same value, -9223372036854775808. The first two work OK but the
> > third throws a overflow error. Why? (Framework v1.0.3705)
> >
> > '-- OK, no overflow
> > Dim i as Long = Long.MinValue
> > Dim k as Long = CLng(-2 ^ 63)
> >
> > '-- Overflow error
> > Dim m as Long = -9223372036854775808
> >
> >
>
> You should add the literal type charecter 'L'.
> Something like -9223372036854775808L

Also doesn't work here (VB 2003). Seems to be a code editor bug.
This works: m = &H8000000000000000


--
Armin

"Ting Liang (MSFT)" <(E-Mail Removed)> schrieb

> > > Dim m as Long = -9223372036854775808


> The 3nd example is obvious to the compiler the number you typed in

thirdond? ;-)

> will never behave as the value as it is entered so it's an error.

Isn't it a Long literal within the valid range for Long literals?



Seem to be still the same code editor programmers like for VB6: Try this in
VB6:

Dim l As Long
l = -2147483648&

It also doesn't work.

*LOL*

--
Armin

hmmmmmm...



I think that the compiler misunderstood.

He is trying to make negative the absolute value, which is an overflow for
Long.

In this case you should use the hexadecimal literal 0x8000000000000000
(&H8000000000000000 in VB)





---

Rafael Pivato





"Brian" <(E-Mail Removed)> escreveu na mensagem
news:(E-Mail Removed)...
> Nope, already tried to append the L.
>
>
> "Rafael Pivato" <(E-Mail Removed)> wrote in message
> news:%(E-Mail Removed)...
> >
> > "Brian" <(E-Mail Removed)> escreveu na mensagem
> > news:(E-Mail Removed)...
> > > Look at the code below. All three variables are initialized to the
same
> > > value, -9223372036854775808. The first two work OK but the third
throws
> a
> > > overflow error. Why? (Framework v1.0.3705)
> > >
> > > '-- OK, no overflow
> > > Dim i as Long = Long.MinValue
> > > Dim k as Long = CLng(-2 ^ 63)
> > >
> > > '-- Overflow error
> > > Dim m as Long = -9223372036854775808
> > >
> > >
> >
> > You should add the literal type charecter 'L'.
> > Something like -9223372036854775808L
> >
> >
> > ----
> > Rafael Pivato
> >
> >
>
>

(E-Mail Removed) (Ting Liang (MSFT)) scripsit:
> The 3nd example is obvious to the compiler the number you typed in will
> never behave as the value as it is entered so it's an error.

I don't understand that. -9223372036854775807 is in the range of the
Long datatype. When specifying it in hex format, it will work.

--
Herfried K. Wagner
MVP · VB Classic, VB.NET
<http://www.mvps.org/dotnet>

Thanks for all of your input. I found out the cause of this problem. If you
want to check it out yourself go to:

http://support.microsoft.com/default...b;en-us;820657



"Armin Zingler" <(E-Mail Removed)> wrote in message
news:e%(E-Mail Removed)...
> "Rafael Pivato" <(E-Mail Removed)> schrieb
> >
> > "Brian" <(E-Mail Removed)> escreveu na mensagem
> > news:(E-Mail Removed)...
> > > Look at the code below. All three variables are initialized to the
> > > same value, -9223372036854775808. The first two work OK but the
> > > third throws a overflow error. Why? (Framework v1.0.3705)
> > >
> > > '-- OK, no overflow
> > > Dim i as Long = Long.MinValue
> > > Dim k as Long = CLng(-2 ^ 63)
> > >
> > > '-- Overflow error
> > > Dim m as Long = -9223372036854775808
> > >
> > >
> >
> > You should add the literal type charecter 'L'.
> > Something like -9223372036854775808L
>
> Also doesn't work here (VB 2003). Seems to be a code editor bug.
> This works: m = &H8000000000000000
>
>
> --
> Armin
>