Unsigned shift right in C# Using Java semantics for negative numbers

Question

I'm trying to port Java code to C# and I'm running into odd bugs related to the unsigned shift right operator >>> normally the code:

long l = (long) ((ulong) number) >> 2;

Would be the equivalent of Java's:

long l = number >>> 2;

However for the case of -2147483648L which you might recognized as Integer.MIN_VALUE this returns a different number than it would in Java since the cast to ulong changes the semantics of the number hence I get a different result.

How would something like this be possible in C#?

I'd like to preserve the code semantics as much as possible since its a pretty complex body of code.

Answer 1

I believe your expression is incorrect when considering C#'s order precedence. Your code I believe is converting your long to ulong, then back to long, then shifting. I'm assuming your intent was to perform the shift on the ulong.

From the C# Specification §7.2.1, Unary (or in your case, the casting operation) takes precedence over the shifting. Thus your code:

long l = (long) ((ulong) number) >> 2;

would be interpreted as:

ulong ulongNumber = (ulong)number;
long longNumber = (long)ulongNumber;
long shiftedlongNumber = longNumber >> 2;

Given number as -2147483648L, this yields 536870912.

By wrapping the conversion and shifting in parenthesis:

long l = (long) (((ulong) number) >> 2);

Produces logic that could be rewritten as:

ulong ulongNumber = (ulong)number;
ulong shiftedulongNumber = ulongNumber >> 2;
long longShiftedNumber = (long)shiftedulongNumber;

Which given number as -2147483648L, this yields 4611686017890516992.

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EDIT: Note that given those ordering rules, there's an extra set of parenthesis in my answer that aren't necessary. The correct expression could be written as:

long l = (long) ((ulong) number >> 2);