什么是 >>>= 来自 C# 中的 Java

Question

I am currently converting a Java Program. And I come by the byte shift operator.

I'd like to know what the >>>= operator means in C#. Is it just >>= ? Does >>= shift the sign in C#?

Answer 1

The >>> syntax in Java is for unsigned right shifts, which is necessary as a concept because Java doesn't have a specific data type for unsigned integers .

However, C# does; in C#, you would just use >> with an unsigned type - so any of ulong , uint , ushort , byte - and it will perform the expected "fill the MSB with zero" behavior, because that is what >> does on unsigned integers, even if the input MSB is set.

If you don't want to change the code to use unsigned types throughout, you can probably use an extension method:

public static int UnsignedRightShift(this int signed, int places)
{
    unchecked // just in case of unusual compiler switches; this is the default
    {
        var unsigned = (uint)signed;
        unsigned >>= places;
        return (int)unsigned;
    }
}
public static long UnsignedRightShift(this long signed, int places)
{
    unchecked // just in case of unusual compiler switches; this is the default
    {
        var unsigned = (ulong)signed;
        unsigned >>= places;
        return (long)unsigned;
    }
}

I've written this long-hand for readability, but the compiler optimizes this pretty well - for example for the int version:

.maxstack 8

ldarg.0
ldarg.1
ldc.i4.s 31
and
shr.un
ret

(the only difference in the long version is that it masks with 63 instead of 31 )

They can be written more tersely as:

public static int UnsignedRightShift(this int signed, int places)
    => unchecked((int)((uint)signed >> places));
public static long UnsignedRightShift(this long signed, int places)
    => unchecked((long)((ulong)signed >> places));