使用转换后的指针值的参考返回值可以吗?
我已经阅读了这个问题。How is return by reference implemented in C#?
关于垃圾收集器的内存移动问题,可以使用下面的代码吗?(Helper.GetReference()和Helper.GetPointer())
class Program
{
static unsafe void Main(string[] args)
{
byte[] bytes = new byte[1024];
ref SomeStruct reference = ref Helper.GetReference<SomeStruct>(bytes);
reference.field1 = 1;
reference.field2 = 2;
SomeStruct* pointer = Helper.GetPointer<SomeStruct>(bytes);
pointer->field1 = 3;
pointer->field2 = 4;
}
}
public static class Helper
{
// Can I use this?
public static unsafe ref T GetReference<T>(byte[] bytes) where T : unmanaged
{
fixed (byte* p1 = bytes)
{
T* p2 = (T*)p1;
return ref *p2;
}
}
// Shouldn't I use it?
public static unsafe T* GetPointer<T>(byte[] bytes) where T : unmanaged
{
fixed (byte* p1 = bytes)
{
return (T*)p1;
}
}
}
public struct SomeStruct
{
public int field1;
public int field2;
}
据我所知,这两个方法都是不安全 ...是的,但是确实可以编译,因为您在unsafe
方法中使用了Helper
并引用了相同的内存,因此安全-net已被删除。
您指向到可以由垃圾收集器(又名数组)移动的一部分内存(托管对象),可能使您留下悬挂指针] >
您需要在fixed
方法中fix
Main
)数组以确保安全(以我的方式看)或您的结构例如
fixed (SomeStruct* fixedPointer = &Helper.GetReference<SomeStruct>(bytes))
{
fixedPointer->field1 = 5;
fixedPointer->field2 = 6;
}
或
fixed (byte* p = bytes) // even though p isn't being used
{
SomeStruct* pointer = Helper.GetPointer<SomeStruct>(bytes);
pointer->field1 = 3;
pointer->field2 = 4;
}