c# - sizeof 根据字段顺序给出不同的结果

Question

我编写了一个小型控制台应用程序来测试 sizeof运算符(operator):

public class Program
{
    public static unsafe void Main(string[] args)
    {
        // Native
        Console.WriteLine("The size of bool is {0}.", sizeof(bool));
        Console.WriteLine("The size of short is {0}.", sizeof(short));
        Console.WriteLine("The size of int is {0}.", sizeof(int));
        Console.WriteLine("The size of long is {0}.", sizeof(long));

        // Custom
        Console.WriteLine("The size of Bool1 is {0}.", sizeof(Bool1));
        Console.WriteLine("The size of Bool2 is {0}.", sizeof(Bool2));
        Console.WriteLine("The size of Bool1Int1Bool1 is {0}.", sizeof(Bool1Int1Bool1));
        Console.WriteLine("The size of Bool2Int1 is {0}.", sizeof(Bool2Int1));
        Console.WriteLine("The size of Bool1Long1 is {0}.", sizeof(Bool1Long1));
        Console.WriteLine("The size of Bool1DateTime1 is {0}.", sizeof(Bool1DateTime1));

        Console.Read();
    }
}

public struct Bool1
{
    private bool b1;
}

public struct Bool2
{
    private bool b1;
    private bool b2;
}

public struct Bool1Int1Bool1
{
    private bool b1;
    private int i1;
    private bool b2;
}

public struct Bool2Int1
{
    private bool b1;
    private bool b2;
    private int i1;
}

public struct Bool1Long1
{
    private bool b1;
    private long l1;
}

public struct Bool1DateTime1
{
    private bool b1;
    private DateTime dt1;
}

给出以下输出:

似乎声明字段的顺序对结构的大小有影响。

我原以为 Bool1Int1Bool1 返回的大小是 6 (1 + 4 + 1)，但它返回的是 12(我想是 4 + 4 + 4？？)!因此，编译器似乎通过按 4 个字节打包 来对齐成员。

如果我在 32 位或 64 位系统上，它会改变什么吗？

第二个问题，对于long类型的测试，这次bool被8字节打包。谁能解释一下？

Answer 1

那是因为编译器对齐成员，因此优化了它们的访问速度，而不是它们的内存占用。

可以添加

[StructLayout(LayoutKind.Sequential, Pack=1)]

在结构定义之前，它应该在 1 个字节的空间内对齐。