我一直在试验 Span<T>
作为 ReadOnlySequence<T>
的一部分和 System.IO.Pipelines。
我目前正在尝试获取 Span<T>
超过 struct
不使用 unsafe
代码,而不复制该 struct
.
我的结构很简单:
[StructLayout(LayoutKind.Sequential, Pack = 1, CharSet = CharSet.Unicode)]
public struct Packet
{
public byte TestByte;
}
方法 1 - 有效 - 但感觉“不安全”
//
// Method 1 - uses Unsafe to get a span over the struct
//
var packet = new Packet();
unsafe
{
var packetSpan = new Span<byte>(&packet, Marshal.SizeOf(packet));
packetSpan[0] = 0xFF; // Set the test byte
Debug.Assert(packet.TestByte == 0xFF, "Error, packetSpan did not update packet.");
// ^^^ Succeeds
packet.TestByte = 0xEE;
Debug.Assert(packetSpan[0] == 0xEE, "Error, packet did not update packetSpan.");
// ^^^ Succeeds
}
方法 2 - 无法按预期工作,因为它需要一份副本
//
// Method 2
//
// This doesn't work as intended because the original packet is actually
// coppied to packet2Array because it's a value type
//
// Coppies the packet to an Array of Packets
// Gets a Span<Packet> of the Array of Packets
// Casts the Span<Packet> as a Span<byte>
//
var packet2 = new Packet();
// create an array and store a copy of packet2 in it
Packet[] packet2Array = new Packet[1];
packet2Array[0] = packet2;
// Get a Span<Packet> of the packet2Array
Span<Packet> packet2SpanPacket = MemoryExtensions.AsSpan<Packet>(packet2Array);
// Cast the Span<Packet> as a Span<byte>
Span<byte> packet2Span = MemoryMarshal.Cast<Packet, byte>(packet2SpanPacket);
packet2Span[0] = 0xFF; // Set the test byte
Debug.Assert(packet2.TestByte == 0xFF, "Error, packet2Span did not update packet2");
// ^^^ fails because packet2 was coppied into the array, and thus packet2 has not changed.
Debug.Assert(packet2Array[0].TestByte == 0xFF, "Error, packet2Span did not update packet2Array[i]");
// ^^^ succeeds
packet2.TestByte = 0xEE;
Debug.Assert(packet2Span[0] == 0xEE, "Error, packet2 did not update in packet2Span");
// ^^^ fails because packet2Span is covering packet2Array which has a copy of packet2
packet2Array[0].TestByte = 0xEE;
Debug.Assert(packet2Span[0] == 0xEE, "Error, packet2 did not update in packet2Span");
// ^^^ succeeds
进一步的研究表明
Span<T>
可以从 byte[]
隐式转换,例如,我可以做Span<byte> packetSpan = new Packet().ToByteArray();
但是我目前的任何 ToByteArray() 实现仍在制作 Packet 结构的副本。
我不能做一些类似的事情:
Span<byte> packetSpan = (byte[])packet;
// ^^ Won't compile
最佳答案
您必须在 unsafe context 中执行此操作因为它是 不安全 根据这个词的真正含义,因为如果你不够小心,你会射中自己的脚。原因如下:
考虑以下代码:
Span<byte> GiveMeSpan()
{
MyLovelyStruct value = new MyLovelyStruct();
unsafe
{
return new Span<byte>(&value, sizeof(MyLovelyStruct));
}
}
MyLovelyStruct
的实例我们在 GiveMeSpan()
中创建的住在方法的call stack而你所做的就是获取它的地址,把它交给 Span<byte>
,并返回 Span<byte>
.一旦一个方法返回,它就会弹出它的 stack frame ,因此您的MyLovelyStruct
的内存生活在将成为免费的,并且可能会被调用者调用的下一个方法回收并破坏它。但这还不是全部,如果您的
MyLovelyStruct
怎么办?生活在这样的对象字段中:class MyLovelyClass
{
private MyLovelyStruct value;
public void Foo()
{
unsafe
{
var span = new Span(&value, sizeof(MyLovelyStruct));
Process(span);
}
}
}
// Declaration
Process(Span<byte> span);
还有一个 GC发生在
Process()
方法正在处理您的 MyLovelyStruct
和 MyLovelyClass
突然在内存中移动(是的,GC 移动内存中的对象,read here)?是的,您的 Span<byte>
指向 MyLovelyStruct
将不再指向新的 MyLovelyStruct
地址并且您的程序已损坏。所以为了安全地包装一个
struct
使用 Span<byte>
或任何其他指针类型,您必须确保:所以
unsafe
关键字是必需的,即使您可以绕过它,您也有责任向代码的读者发出警告。
关于c# - 在结构上获取 Span<byte> 而不复制结构,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59596364/