这是原始声明:
[StructLayout(LayoutKind.Explicit, Size = 16)]
public unsafe struct X
{
[FieldOffset(0)] public ushort a;
[FieldOffset(2)] public fixed byte b[14];
};
我想将struct
设为只读,但我不知道应该如何编写数组的getter。我能想出的唯一解决方案是getter 方法:
[StructLayout(LayoutKind.Explicit, Size = 16)]
public unsafe struct X
{
[FieldOffset(0)] private ushort a;
[FieldOffset(2)] private fixed byte b[14];
public ushort A { get { return a; } }
public byte B(int i) { fixed (byte* p = b) { return p[i]; } }
};
是否可以为b而不是getter 方法编写getter 属性?
===更新===
我还想处理有多个数组字段的情况。例如:
[StructLayout(LayoutKind.Explicit, Size = 24)]
public unsafe struct Y
{
[FieldOffset(0)] private ushort a;
[FieldOffset(2)] private fixed byte b[14];
[FieldOffset(16)] private fixed byte c[8];
public ushort A { get { return a; } }
public byte B(int i) { fixed (byte* p = b) { return p[i]; } }
public byte C(int i) { fixed (byte* p = c) { return p[i]; } }
};
是否可以为b和c而不是getter 方法编写getter 属性?我想写y.B[i]
和y.C[i]
而不是y.B(i)
和y.C(i)
。
答案 0 :(得分:3)
您可以使用索引器属性来访问特定索引处的数组。不幸的是,它需要在结构本身而不是B上定义,但这应该提供您正在寻找的内容:
[StructLayout(LayoutKind.Explicit, Size = 16)]
public unsafe struct X
{
[FieldOffset(0)]
private ushort a;
[FieldOffset(2)]
private fixed byte b[14];
public ushort A { get { return a; } }
public byte this [int i]
{
get
{
byte b1;
fixed (byte* b2 = b)
{
b1 = b2[i];
}
return b1;
}
}
};
答案 1 :(得分:1)
我找到了一个解决方案,它基于indexed properties:
的想法[StructLayout(LayoutKind.Explicit, Size = 14)]
public struct YB
{
[FieldOffset(0)] private fixed byte b[14];
public byte this[int i] { get { fixed (byte* p = b) { return p[i]; } } }
}
[StructLayout(LayoutKind.Explicit, Size = 8)]
public struct YC
{
[FieldOffset(0)] private fixed byte c[8];
public byte this[int i] { get { fixed (byte* p = c) { return p[i]; } } }
}
[StructLayout(LayoutKind.Explicit, Size = 24)]
public unsafe struct Y
{
[FieldOffset(0)] private ushort a;
[FieldOffset(2)] private YB b;
[FieldOffset(16)] private YC c;
public ushort A { get { return a; } }
public YB B { get { return b; } }
public YC C { get { return c; } }
};