Question

这是“软问题”的范畴,但基本上,Im试图采取这种模式,利用隐性转换成特:

byte x = 0;
x |= 2;       // implicit conversion of "x | 2" to byte here

http://www.em>removingbits:

x &= ~2;      // compile error

我可以谈得最短的是:

x &= unchecked((byte) ~2);

(在这一点上,我开始认真考虑仅写以下文字:<代码>x &=253;/ ~2<>/code>,改写,或仅考虑精老的明文:x = (byte)(x &~2);

我没有一个较短的路?

Answer 1

如何做到这一点:

{
   byte x = 0;
   x |= 2;
   x &= 0xFF & ~2;
}

解释: 这里有两个问题。第一,固定运营商:

C# Spec 4.1.5:

The integral-type unary and binary operators always operate with signed 32-bit precision, unsigned 32-bit precision, signed 64-bit precision, or unsigned 64-bit precision:

For the unary + and ~ operators, the operand is converted to type T, where T is the first of int, uint, long, and ulong that can fully represent all possible values of the operand. The operation is then performed using the precision of type T, and the type of the result is T.

一旦您申请独立经营人,结果总是在最小的某个类型。从那以后,你想要暗中地转换成秘密。

第二,。

A constant expression of type int can be converted to sbyte, byte, short, ushort, uint, or ulong, provided the value of the constant expression is within the range of the destination type.

因此,大约2,总是暗中。不能不言而喻地加以转换,因为它不属于范围。如果你限制在幅度之内,你可以暗中转换。

Answer 2

你们可以这样说:

[Flags]
enum SomeType : byte {
  Value = 2,
}

public void Method () {
  SomeType t = 0;
  t |= SomeType.Value;
  t &= ~SomeType.Value;
}

Answer 3

使用习俗类型

    public struct Bits
    {
        byte x;
        public Bits(int x) { this.x = (byte)x; }
        public Bits(byte x) { this.x = x; }            
        public Bits(Bits x) { this.x = x.x; }

        public static implicit operator byte(Bits x) { return x.x; }
        public static implicit operator Bits(byte x) { return new Bits(x); }
        public static implicit operator Bits(int x) { return new Bits(x); }
    }

    static void Main(string[] args)
    {
        Bits x = 0;
        x |= 2;
        x &= ~2;
    }

此外,你还可以用逐级推展法确定借方。

    public static byte Set(this byte x, byte v)
    {
        return (byte)(x | v);
    }
    public static byte Unset(this byte x, byte v)
    {
        return (byte)(x & ~v);
    }
    static void Main(string[] args)
    {
        byte x = 0;
        x = x.Set(2);
        x = x.Unset(2);
    }

Answer 4

byte x = 6;
Console.WriteLine(x & ~(byte)2);

我的著作,第4版

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