为什么将 ref struct 传递给方法实际上传递了 ref struct 的副本？

当我试图传递一个看起来像的 ref 结构时，说

public ref struct BufferWriter
{
    readonly Buffer* _pb;
    public int WrittenBytes;
    //...
    public bool Append(ReadOnlySpan<byte> buffer)
    {
        //...
        WrittenBytes += buffer.Length;
        //...
    }
    
}

对于方法（Write100U8Chars(BufferWriter, Object)在下面的示例中），结构实际上不是通过引用传递的，而是通过值传递的：

protected override bool Serialize(object obj)
{
    //...
    fixed (Buffer* pb = &doc_body)
    {
        using (BufferWriter writer = new BufferWriter(pb))
        {
            writer.Append("UTF8 str"U8);
            // pb->Bytes is now 8
            // writer.WrittenBytes is now also 8
            Write100U8Chars(writer, obj);
            // pb->Bytes is now 108
            // But writer.WrittenBytes is still 8!
        }
    }
    //...
}

// A method that always calls BufferWriter.Append passing 100 UTF-8 chars (aka bytes) 
protected abstract bool Write100U8Chars(BufferWriter writer, object obj);

由于using块是一种只读上下文，正如我所想的那样，强制制作“防御性副本”，我试图删除块using/指令：

fixed (Buffer* pb = &doc_body)
{
    BufferWriter writer = new BufferWriter(pb)
    writer.Append("UTF8 str"U8);
    // writer.WrittenBytes is now 8
    Write100U8Chars(writer, obj);
    // No use of removing using: writer.WrittenBytes is still 8
}

做BufferWriter一个ref readonly struct没有区别。

将局部变量传递给带有参数的writer方法仍然没有区别。Write100U8Chars(BufferWriter, Object)in

如何ref struct通过引用而不是像普通非引用结构那样通过值将 a 传递给抽象方法？