在 VB.NET 到 C# 中显示不同结果的二进制移位

Question

I'm really confused as to why the following code gives me different results when run in VB.NET as opposed to C#. I've read that there are some binary shift differences between the two languages but I can't work out what I need to do to make VB.NET show the same answer for 'pos' as C# does.

Imports System.Threading

Public Class Form1
    Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click

        MultiplyH({168, 238, 95, 83, 235, 11, 228, 190, 8, 58, 0, 0, 0, 0, 0, 0})


    End Sub

    Public Sub MultiplyH(ByVal x As Byte())

        Dim pos As Integer = x(15) << 1
        Dim i = 14
        While i >= 0
            pos = x(i) << 1
            Interlocked.Decrement(i)
        End While
    End Sub
End Class

The first eight answers for 'pos' are all the same (to be expected for the zeroes) but once i = 6, VB.NET shows pos = 124 whereas C# has pos = 380. Can anyone help me so that pos = 380 in VB.NET please?

The C# code, btw, is:

int pos = x[15] << 1;
for (int i = 14; i >= 0; --i)
{
                pos = x[i] << 1;
}

Answer 1

From the documentation for the C# << operator :

Because the shift operators are defined only for the int, uint, long, and ulong types, the result of an operation always contains at least 32 bits. If the left-hand operand is of another integral type (sbyte, byte, short, ushort, or char), its value is converted to the int type

[Emphasis mine.]

From the documentation for the VB.NET << operator :

The result always has the same data type as that of the expression being shifted.

To get the same result in VB.NET as in C#, you need to use:

pos = CInt(x(i)) << 1