c# - 使用 Expression> 使值类型表现为引用类型

Question

我们知道 int 是一种值类型，因此以下内容是有道理的:

int x = 3;
int y = x;
y = 5;
Console.WriteLine(x); //says 3. 

现在，由于缺少更好的术语，我们想要将两个变量“链接”到相同的内存位置，这是一些代码。

int x = 3;
var y = MagicUtilClass.linkVariable(() => x);
y.Value = 5;
Console.WriteLine(x) //says 5.

问题是:方法 linkVariable 是什么样子的？它的返回类型是什么样的？

尽管我将帖子命名为使值类型表现为引用类型，但上述 linkVariable 方法也适用于引用类型......，即，

Person x = new Person { Name = "Foo" };
var y = MagicUtilClass.linkVariable(() => x);
y.Value = new Person { Name = "Bar" };
Console.WriteLine(x.Name) //says Bar.

我不确定如何在 C# 中实现这一点(顺便说一句，不允许使用不安全的代码)？

欣赏创意。谢谢。

Answer 1

这是一个完整的解决方案:

// Credits to digEmAll for the following code
public delegate void Setter<T>(T newValue);
public delegate T Getter<T>();
public class MagicPointer<T>
{
    private Getter<T> getter;
    private Setter<T> setter;

    public T Value
    {
        get { return getter(); }
        set { setter(value); }
    }

    public MagicPointer(Getter<T> getter, Setter<T> setter)
    {
        this.getter = getter;
        this.setter = setter;
    }
}

// Code starting from here is mine
public static class MagicUtilClass
{
    public static MagicPointer<T> LinkVariable<T>(Expression<Func<T>> expression)
    {
        var memberExpr = expression.Body as MemberExpression;
        if (memberExpr == null)
            throw new InvalidOperationException("The body of the expression is expected to be a member-access expression.");
        var field = memberExpr.Member as FieldInfo;
        if (field == null)
            throw new InvalidOperationException("The body of the expression is expected to be a member-access expression that accesses a field.");
        var constant = memberExpr.Expression as ConstantExpression;
        if (constant == null)
            throw new InvalidOperationException("The body of the expression is expected to be a member-access expression that accesses a field on a constant expression.");
        return new MagicPointer<T>(() => (T) field.GetValue(constant.Value),
                                   x => field.SetValue(constant.Value, x));
    }
}

用法:

int x = 47;
var magic = MagicUtilClass.LinkVariable(() => x);
magic.Value = 48;
Console.WriteLine(x);  // Outputs 48

要了解此解决方案为何有效，您需要知道每当您在 lambda 表达式中使用变量时，编译器都会相当大地转换您的代码(无论该 lambda 表达式是成为委托(delegate)还是表达式树)。它实际上生成了一个包含字段的新类。变量 x 被删除并替换为该字段。用法示例将如下所示:

CompilerGeneratedClass1 locals = new CompilerGeneratedClass1();
locals.x = 47;
var magic = MagicUtilClass.LinkVariable(() => locals.x);
// etc.

代码检索的“字段”是包含x 的字段，它检索的“常量”是locals 实例。