C#中的位字段
C#中的位字段
提问于 2008-08-18 11:19:28
我有一个结构，我需要填充和写入磁盘(实际上有几个)。
下面是一个例子：
byte-6
bit0 - original_or_copy
bit1 - copyright
bit2 - data_alignment_indicator
bit3 - PES_priority
bit4-bit5 - PES_scrambling control.
bit6-bit7 - reserved 在C语言中，我可能会做类似以下的事情：
struct PESHeader {
unsigned reserved:2;
unsigned scrambling_control:2;
unsigned priority:1;
unsigned data_alignment_indicator:1;
unsigned copyright:1;
unsigned original_or_copy:1;
};有没有办法在C#中使用结构解引用点运算符来访问这些位？
对于几个结构，我只需要在一个存取器函数中进行位移位。
我有大量的结构要以这种方式处理，所以我正在寻找更容易阅读和更快编写的东西。
回答 12
Stack Overflow用户
回答已采纳
发布于 2008-08-18 13:31:31
我可能会拼凑一些使用属性的东西，然后是一个转换类，将适当的属性结构转换为位域原语。就像..。
using System;
namespace BitfieldTest
{
[global::System.AttributeUsage(AttributeTargets.Field, AllowMultiple = false)]
sealed class BitfieldLengthAttribute : Attribute
{
uint length;
public BitfieldLengthAttribute(uint length)
{
this.length = length;
}
public uint Length { get { return length; } }
}
static class PrimitiveConversion
{
public static long ToLong<T>(T t) where T : struct
{
long r = 0;
int offset = 0;
// For every field suitably attributed with a BitfieldLength
foreach (System.Reflection.FieldInfo f in t.GetType().GetFields())
{
object[] attrs = f.GetCustomAttributes(typeof(BitfieldLengthAttribute), false);
if (attrs.Length == 1)
{
uint fieldLength = ((BitfieldLengthAttribute)attrs[0]).Length;
// Calculate a bitmask of the desired length
long mask = 0;
for (int i = 0; i < fieldLength; i++)
mask |= 1 << i;
r |= ((UInt32)f.GetValue(t) & mask) << offset;
offset += (int)fieldLength;
}
}
return r;
}
}
struct PESHeader
{
[BitfieldLength(2)]
public uint reserved;
[BitfieldLength(2)]
public uint scrambling_control;
[BitfieldLength(1)]
public uint priority;
[BitfieldLength(1)]
public uint data_alignment_indicator;
[BitfieldLength(1)]
public uint copyright;
[BitfieldLength(1)]
public uint original_or_copy;
};
public class MainClass
{
public static void Main(string[] args)
{
PESHeader p = new PESHeader();
p.reserved = 3;
p.scrambling_control = 2;
p.data_alignment_indicator = 1;
long l = PrimitiveConversion.ToLong(p);
for (int i = 63; i >= 0; i--)
{
Console.Write( ((l & (1l << i)) > 0) ? "1" : "0");
}
Console.WriteLine();
return;
}
}
}它会生成预期的...000101011。当然，它需要更多的错误检查和稍微清醒的类型，但它的概念(我认为)是健全的，可重用的，并且可以让你敲出十几个容易维护的结构。
adamw
Stack Overflow用户
发布于 2008-08-18 12:25:52
通过使用枚举，您可以做到这一点，但看起来会很笨拙。
[Flags]
public enum PESHeaderFlags
{
IsCopy = 1, // implied that if not present, then it is an original
IsCopyrighted = 2,
IsDataAligned = 4,
Priority = 8,
ScramblingControlType1 = 0,
ScramblingControlType2 = 16,
ScramblingControlType3 = 32,
ScramblingControlType4 = 16+32,
ScramblingControlFlags = ScramblingControlType1 | ScramblingControlType2 | ... ype4
etc.
}Stack Overflow用户
发布于 2008-08-18 11:32:53
[StructLayout(LayoutKind.Explicit, Size=1, CharSet=CharSet.Ansi)]
public struct Foo
{ [FieldOffset(0)]public byte original_or_copy;
[FieldOffset(0)]public byte copyright;
[FieldOffset(0)]public byte data_alignment_indicator;
[FieldOffset(0)]public byte PES_priority;
[FieldOffset(0)]public byte PES_scrambling_control;
[FieldOffset(0)]public byte reserved;
}这实际上是一个联合，但您可以将其用作位字段--您只需意识到每个字段的位应该在字节中的什么位置。实用函数和/或常量可以提供帮助。
const byte _original_or_copy = 1;
const byte _copyright = 2;
//bool ooo = foo.original_or_copy();
static bool original_or_copy(this Foo foo)
{ return (foo.original_or_copy & _original_or_copy) == original_or_copy;
} 还有一个LayoutKind.Sequential，它可以让你用C语言来做。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/14464
复制相关文章
相似问题