将大端字节集合编组到结构中以提取值 [英] Marshalling a big-endian byte collection into a struct in order to pull out values
问题描述
有一个关于 阅读 C 的有见地的问题来自字节数组的 C# 中的/C++ 数据结构，但我无法让代码适用于我的大端(网络字节顺序)字节集合.(请注意，我真正的结构不止一个字段.)有没有办法将字节编组为结构的大端版本，然后提取框架字节序中的值(主机的字节序)，通常是小端)?
There is an insightful question about reading a C/C++ data structure in C# from a byte array, but I cannot get the code to work for my collection of big-endian (network byte order) bytes. ( Note that my real struct has more than just one field.) Is there a way to marshal the bytes into a big-endian version of the structure and then pull out the values in the endianness of the framework (that of the host, which is usually little-endian)?
(请注意，反转字节数组将不起作用 - 每个值的字节必须反转，这不会为您提供与反转所有字节相同的集合.)
(Note, reversing the array of bytes will not work - each value's bytes must be reversed, which does not give you the same collection as reversing all of the bytes.)
这应该总结我在寻找什么(LE=LittleEndian，BE=BigEndian):
This should summarize what I'm looking for (LE=LittleEndian, BE=BigEndian):
void Main()
{
var leBytes = new byte[] {1, 0, 2, 0};
var beBytes = new byte[] {0, 1, 0, 2};
Foo fooLe = ByteArrayToStructure<Foo>(leBytes);
Foo fooBe = ByteArrayToStructureBigEndian<Foo>(beBytes);
Assert.AreEqual(fooLe, fooBe);
}
[StructLayout(LayoutKind.Explicit, Size=4)]
public struct Foo {
[FieldOffset(0)]
public ushort firstUshort;
[FieldOffset(2)]
public ushort secondUshort;
}
T ByteArrayToStructure<T>(byte[] bytes) where T: struct
{
GCHandle handle = GCHandle.Alloc(bytes, GCHandleType.Pinned);
T stuff = (T)Marshal.PtrToStructure(handle.AddrOfPinnedObject(),typeof(T));
handle.Free();
return stuff;
}
T ByteArrayToStructureBigEndian<T>(byte[] bytes) where T: struct
{
???
}
其他有用的链接:
推荐答案
正如我在@weismat 的回答中提到的，有一种简单的方法可以实现大端结构.它涉及双重反转:将原始字节完全反转，然后结构本身是原始(大端)数据格式的反转.
As alluded to in my comment on @weismat's answer, there is an easy way to achieve big-endian structuring. It involves a double-reversal: the original bytes are reversed entirely, then the struct itself is the reversal of the original (big-endian) data format.
Main 中的fooLe 和fooBe 对于所有字段将具有相同的值.(通常情况下，little-endian 结构和字节当然不会出现，但这清楚地表明了字节顺序之间的关系.)
The fooLe and fooBe in Main will have the same values for all fields. (Normally, the little-endian struct and bytes wouldn't be present, of course, but this clearly shows the relationship between the byte orders.)
注意:请参阅更新代码，包括如何从结构中取回字节.
NOTE: See updated code including how to get bytes back out of the struct.
public void Main()
{
var beBytes = new byte[] {
0x80,
0x80,0,
0x80,0,
0x80,0,0,0,
0x80,0,0,0,
0x80,0,0,0,0,0,0,0,
0x80,0,0,0,0,0,0,0,
0x3F,0X80,0,0, // float of 1 (see http://en.wikipedia.org/wiki/Endianness#Floating-point_and_endianness)
0x3F,0xF0,0,0,0,0,0,0, // double of 1
0,0,0,0x67,0x6E,0x69,0x74,0x73,0x65,0x54 // Testing
};
var leBytes = new byte[] {
0x80,
0,0x80,
0,0x80,
0,0,0,0x80,
0,0,0,0x80,
0,0,0,0,0,0,0,0x80,
0,0,0,0,0,0,0,0x80,
0,0,0x80,0x3F, // float of 1
0,0,0,0,0,0,0xF0,0x3F, // double of 1
0x54,0x65,0x73,0x74,0x69,0x6E,0x67,0,0,0 // Testing
};
Foo fooLe = ByteArrayToStructure<Foo>(leBytes).Dump("LE");
FooReversed fooBe = ByteArrayToStructure<FooReversed>(beBytes.Reverse().ToArray()).Dump("BE");
}
[StructLayout(LayoutKind.Sequential, Pack = 1)]
public struct Foo {
public byte b1;
public short s;
public ushort S;
public int i;
public uint I;
public long l;
public ulong L;
public float f;
public double d;
[MarshalAs(UnmanagedType.ByValTStr, SizeConst = 10)]
public string MyString;
}
[StructLayout(LayoutKind.Sequential, Pack = 1)]
public struct FooReversed {
[MarshalAs(UnmanagedType.ByValTStr, SizeConst = 10)]
public string MyString;
public double d;
public float f;
public ulong L;
public long l;
public uint I;
public int i;
public ushort S;
public short s;
public byte b1;
}
T ByteArrayToStructure<T>(byte[] bytes) where T: struct
{
GCHandle handle = GCHandle.Alloc(bytes, GCHandleType.Pinned);
T stuff = (T)Marshal.PtrToStructure(handle.AddrOfPinnedObject(),typeof(T));
handle.Free();
return stuff;
}
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