如何将递归结构编组为c sharp?

时间:2009-07-19 16:05:04

标签: c# marshalling recursive-datastructures

我有一个非管理的结构,我想整理一下,看起来基本上是这样的:

struct MyStruct{  
    /* ... some stuff ... */
    int numChilds;  
    MyStruct *childs;
}

我相信我必须写一个自定义编组,但我不确定如何继续。

2 个答案:

答案 0 :(得分:3)

当我不需要直接索引孩子时,我喜欢使用这样的设置:

struct MyStruct
{
    /* ... some stuff ... */
    int numChilds;
    IntPtr childData;

    public IEnumerable<MyStruct> Children
    {
        get
        {
            int elementSize = Marshal.SizeOf(typeof(MyStruct));
            for (int i = 0; i < this.numChilds; i++)
            {
                IntPtr data = new IntPtr(this.childData.ToInt64() + elementSize * i);
                MyStruct child = (MyStruct)Marshal.PtrToStructure(data, typeof(MyStruct));
                yield return child;
            }
        }
    }
}

如果你需要直接索引孩子,最简单的方法是创建一个方法GetChild(如下所示)。更难的方法是创建一个实现IList<MyStruct>的帮助器/包装器类。将从Children属性返回一个实例,其内部将通过调用GetChild方法来工作。如果需要,这可以留给读者练习。

public MyStruct GetChild(int index)
{
    if (index < 0)
        throw new ArgumentOutOfRangeException("index", "The index must be >= 0.");
    if (index >= this.numChilds)
        throw new ArgumentException("The index must be less than the number of children", "index");

    int elementSize = Marshal.SizeOf(typeof(MyStruct));
    IntPtr data = new IntPtr(childData.ToInt64() + elementSize * index);
    MyStruct child = (MyStruct)Marshal.PtrToStructure(data, typeof(MyStruct));
    return child;
}

答案 1 :(得分:0)

如果你只想将它传递给一些非托管函数,你可以使用不安全的代码和stackalloc / fix一个数组来获取一个指向对象数组的指针。

        unsafe struct Foo
        {
            public int value;
            public int fooCount;
            public Foo* foos;
        }

        [DllImport("dll_natv.dll")]
        static extern void PrintFoos(Foo f);

        public unsafe static void Main()
        {
            Foo* foos = stackalloc Foo[10];

            for (int i = 0; i < 10; ++i)
                foos[i].value = i;

            Foo mainFoo = new Foo();
            mainFoo.fooCount = 10;
            mainFoo.value = 100;
            mainFoo.foos = foos;

            PrintFoos(mainFoo);


        }