等效于 C# 中的 Java 三重移位运算符 (>>>)？
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Equivalent of Java triple shift operator (>>>) in C#?
提问by Nikolaos
What is the equivalent (in C#) of Java's >>>operator?
Java>>>运算符的等价物（在 C# 中）是什么？
(Just to clarify, I'm not referring to the >>and <<operators.)
（只是为了澄清，我不是指>>和<<运营商。）
采纳答案by Lucero
In C#, you can use unsigned integer types, and then the <<and >>do what you expect. The MSDN documentation on shift operatorsgives you the details.
在C＃中，你可以使用无符号整数类型，然后<<和>>你的期望。有关班次运算符的MSDN 文档为您提供了详细信息。
Since Java doesn't support unsigned integers (apart from char), this additional operator became necessary.
由于 Java 不支持无符号整数（除了char），这个额外的运算符就变得必要了。
回答by Matt
Java doesn't have an unsigned left shift (<<<), but either way, you can just cast to uintand shfit from there.
Java 没有无符号左移 ( <<<)，但无论哪种方式，您都可以uint从那里强制转换和 shfit。
E.g.
例如
(int)((uint)foo >> 2); // temporarily cast to uint, shift, then cast back to int
回答by KeithC
Upon reading this, I hope my conclusion of use as follows is correct. If not, insights appreciated.
阅读本文后，我希望我的以下使用结论是正确的。如果没有，见解赞赏。
Java
爪哇
i >>>= 1;
C#:
C＃：
i = (int)((uint)i >> 1);
回答by Sebastien Lebreton
n >>> s in Java is equivalent to TripleShift(n,s) where:
Java 中的 n >>> s 等价于 TripleShift(n,s) 其中：
private static long TripleShift(long n, int s)
{
if (n >= 0)
return n >> s;
return (n >> s) + (2 << ~s);
}
回答by Musakkhir Sayyed
There is no >>> operator in C#. But you can convert your value like int,long,Int16,Int32,Int64 to unsigned uint, ulong, UInt16,UInt32,UInt64 etc.
C# 中没有 >>> 运算符。但是您可以将 int,long,Int16,Int32,Int64 等值转换为无符号 uint、ulong、UInt16、UInt32、UInt64 等。
Here is the example.
这是示例。
private long getUnsignedRightShift(long value,int s)
{
return (long)((ulong)value >> s);
}
回答by Charles Okwuagwu
For my VB.Netfolks
对于我的VB.Net乡亲
The suggested answers above will give you overflow exceptions with Option Strict ON
上面的建议答案会给你溢出异常 Option Strict ON
Try this for example -100 >>> 2with above solutions:
例如，-100 >>> 2使用上述解决方案尝试此操作：
The following code works always for >>>
以下代码始终适用于 >>>
Function RShift3(ByVal a As Long, ByVal n As Integer) As Long
If a >= 0 Then
Return a >> n
Else
Return (a >> n) + (2 << (Not n))
End If
End Function
