VB.NET code returning different result than C# code

Question

The following code in C# gives result "228452386"

UInt32 a;

int k = 0;

a = 0x9E3779B9;

a += (UInt32)(url[k + 0] + (url[k + 1] << 8) + (url[k + 2] << 16) + (url[k + 3] << 24));

After above code is ran, "a" contains "228452386".

But the following same code in VB.NET results in "Arithmetic operation resulted in an overflow". Basically the last statement is returning value "1868983913", so the runtime error is generated.

Dim a As UInt32

Dim k As Integer = 0

a = &H9E3779B9UI

a += CUInt(AscW(url(k + 0)) + (AscW(url(k + 1)) << 8) + (AscW(url(k + 2)) << 16) + (AscW(url(k + 3)) << 24))

Please note the variable "url" in the above code could be any string and it is same for both codes.

When I run the following statements in both C# and VB.NET than they both return same value

C#

(UInt32)(url[k + 0] + (url[k + 1] << 8) + (url[k + 2] << 16) + (url[k + 3] << 24))

VB.NET

CUInt(AscW(url(k + 0)) + (AscW(url(k + 1)) << 8) + (AscW(url(k + 2)) << 16) + (AscW(url(k + 3)) << 24))

Both statements return the value "1868983913" for "url" "info:microsoft.com". But when I run the complete statement with a += ......... then VB.NET gives an error while C# returns the value "228452386".

You should probably use Asc instead of AscW. See msdn.microsoft.com/en-us/library/zew1e4wc(v=vs.80).aspx — Sani Huttunen
– Sani Huttunen, Commented Apr 20, 2012 at 8:04
To remove the error you need to disable integer overflow checks in the Advanced Compiler Options dialog. — Sani Huttunen
– Sani Huttunen, Commented Apr 20, 2012 at 8:13
You have started 3 questions about this. stackoverflow.com/questions/10241027/…, stackoverflow.com/questions/10242355/…, stackoverflow.com/questions/9333540/… — Dejan Janjušević
– Dejan Janjušević, Commented Apr 20, 2012 at 8:20
@DejanJanjušević They are all related to different problems. — Ali
– Ali, Commented Apr 20, 2012 at 8:58

Arjan Einbu · Accepted Answer · 2012-04-20 10:25:02Z

2

You have an overflow, because the result of adding the two numbers you're adding are larger than what fits in an UInt.

The resulting number is a 33 bit number, UInts fit only 32 bits.

In C#, the most significant bit is just chopped of, keeping the 32 bits that fit inside the UInt. (This new value is not the actual result of adding your numbers.)
In VB.NET you get an exception stating that the number doesn't fit.

The VB.NET compiler (or VS) allows you to turn of this overflow checking, so that VB.NET will behave like C# in this case.

answered Apr 20, 2012 at 10:25

Arjan Einbu

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Arjan Einbu · Accepted Answer · 2012-04-20 10:19:38Z

1

VB.NET automatically inserts tests for overflow, while C# doesn't. The equivalent C# code would be:

UInt32 a;
int k = 0;
a = 0x9E3779B9;
checked{
    a += (UInt32)(url[k + 0] + (url[k + 1] << 8) + (url[k + 2] << 16) + (url[k + 3] << 24));
}

EDIT: Changed code sample to comply with edited question.

edited Apr 20, 2012 at 10:19

answered Apr 20, 2012 at 8:02

Arjan Einbu

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3 Comments

Ali Over a year ago

Basically I want equivalent code for VB.Net based on C# code.

Arjan Einbu Over a year ago

I didn't find the equivalent "unchecked" keyword for VB. (Don't know if it has.) But you can turn of overflow checks for your VB.NET application, as stated here: stackoverflow.com/a/4515123/19594

Ali Over a year ago

Please check my EDIT above as I have added more info.

Dejan Janjušević · Accepted Answer · 2012-04-20 10:48:51Z

0

Use the following VB.NET code:

    a += CUInt(AscW(url(k + 0))) + CUInt((AscW(url(k + 1)) << 8)) + CUInt((AscW(url(k + 2)) << 16)) + CUInt((AscW(url(k + 3)) << 24))

AscW returns an Integer (signed) and what you need is unsigned.

EDIT

This works:

    Dim a As Int64
    Dim result As UInt32

    Dim k As Integer = 0

    Dim url As String = "info:microsoft.com"

    a = &H9E3779B9UI

    Dim a1 As UInt32 = 0
    Dim a2 As UInt32 = 0
    Dim a3 As UInt32 = 0
    Dim a4 As UInt32 = 0

    a1 = AscW(url(k + 0))
    a2 = AscW(url(k + 1)) << 8
    a3 = AscW(url(k + 2)) << 16
    a4 = AscW(url(k + 3)) << 24

    a = (a + a1 + a2 + a3 + a4) And &HFFFFFFF
    result = Convert.ToUInt32(a)

edited Apr 20, 2012 at 10:48

answered Apr 20, 2012 at 8:13

Dejan Janjušević

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3 Comments

Ali Over a year ago

I got same error when I used the above statement because it is also giving value "1868983913" as stated in my original problem above.

Dejan Janjušević Over a year ago

It returned the same value on my PC, which string did you use for url?

Ali Over a year ago

Please check my EDIT above as I have added more info.

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