Lazy <T>如何解决需要new()约束的问题？

Question

示例1(不编译):

void Main()
{
    var c = new C<D>();
    c.M.F();
}

class C<T>
{
    T _m = null;
    public T M { get {
        if(_m == null) _m = new T();
        return _m; 
    } }
}

class D
{
    public void F() { Console.WriteLine ("i was created"); }
}

结果:

无法创建变量类型"T"的实例,因为它没有new()约束

例2(作品):

void Main()
{
    var c = new C<D>();
    c.M.F();
}

class C<T>
{
    Lazy<T> _m = new Lazy<T>();
    public T M { get { return _m.Value; } }
}

class D
{
    public void F() { Console.WriteLine ("i was created"); }
}

结果:

i was created

Answer 1

如果您深入研究源代码,您将看到Lazy<T>最终使用Activator:

return new Lazy<T>.Boxed((T)Activator.CreateInstance(typeof(T)));

这只是使用反射的捷径.由于它不通过实际泛型类型参数(new T())实例化类型,而是通过反射调用构造函数,因此不需要where T : new()约束.