到目前为止,我刚刚理所当然地认为Marshal.SizeOf是计算非托管堆上blittable结构的内存大小的正确方法(这似乎是SO上的共识,几乎是网络上的其他地方).

但在阅读了一些针对Marshal.SizeOf的警告之后(this article之后“但是有问题……”)我试了一下,现在我完全糊涂了：

public struct TestStruct
{
    public char x;
    public char y;
}

class Program
{
    public static unsafe void Main(string[] args)
    {
        TestStruct s;
        s.x = (char)0xABCD;
        s.y = (char)0x1234;

        // this results in size 4 (two Unicode characters)
        Console.WriteLine(sizeof(TestStruct));

        TestStruct* ps = &s;

        // shows how the struct is seen from the managed side... okay!      
        Console.WriteLine((int)s.x);
        Console.WriteLine((int)s.y);

        // shows the same as before (meaning that -> is based on 
        // the same memory layout as in the managed case?)... okay!
        Console.WriteLine((int)ps->x);
        Console.WriteLine((int)ps->y);

        // let's try the same on the unmanaged heap
        int marshalSize = Marshal.SizeOf(typeof(TestStruct));
        // this results in size 2 (two single byte characters)
        Console.WriteLine(marshalSize);

        TestStruct* ps2 = (TestStruct*)Marshal.AllocHGlobal(marshalSize);

        // hmmm, put to 16 bit numbers into only 2 allocated 
        // bytes, this must surely fail...
        ps2->x = (char)0xABCD;
        ps2->y = (char)0x1234;

        // huh??? same result as before, storing two 16bit values in 
        // only two bytes??? next will be a perpetuum mobile...
        // at least I'd expect an access violation
        Console.WriteLine((int)ps2->x);
        Console.WriteLine((int)ps2->y);

        Console.Write("Press any key to continue . . . ");
        Console.ReadKey(true);
    }
}

这里出了什么问题？字段解除引用运算符’ – >’的内存布局假设？是’ – >’甚至是解决非托管结构的合适操作符？或Marshal.SizeOf错误的大小操作符为非托管结构？

我发现没有什么能用我理解的语言解释这一点.除了“…结构布局是不可发现的……”和“……在大多数情况下……”这种愚蠢的东西.

解决方法:

我认为你仍然没有回答的一个问题是你的特殊情况：

&ps2->x
0x02ca4370  <------
    *&ps2->x: 0xabcd 'ꯍ'
&ps2->y
0x02ca4372  <-------
    *&ps2->y: 0x1234 'ሴ'

您正在写入(可能)未分配的内存并从中读取.由于您所在的存储区域,因此未检测到.

这将重现预期的行为(至少在我的系统上,YMMV)：

  TestStruct* ps2 = (TestStruct*)Marshal.AllocHGlobal(marshalSize*10000);

  // hmmm, put to 16 bit numbers into only 2 allocated 
  // bytes, this must surely fail...
  for (int i = 0; i < 10000; i++)
  {
    ps2->x = (char)0xABCD;
    ps2->y = (char)0x1234;
    ps2++;
  }