试图使 Feature 泛型然后突然编译器说
Trying to make Feature generic and then suddenly compiler said
运算符'?'不能应用于T"类型的操作数
这里是代码
public abstract class Feature<T>
{
public T Value
{
get { return GetValue?.Invoke(); } // here is error
set { SetValue?.Invoke(value); }
}
public Func<T> GetValue { get; set; }
public Action<T> SetValue { get; set; }
}
可以改用这段代码
get
{
if (GetValue != null)
return GetValue();
return default(T);
}
但我想知道如何修复那个漂亮的 C# 6.0 单线.
But I am wondering how to fix that nice C# 6.0 one-liner.
由于不是所有的东西都可以为 null，你必须缩小 T 为可以为空的东西(又名对象).结构不能为空，枚举也不能.
Since not everything can be null, you have to narrow down T to be something nullable (aka an object). Structs can't be null, and neither can enums.
在 class 上添加 where 确实可以解决问题:
Adding a where on class does fix the issue:
public abstract class Feature<T> where T : class
那么为什么它不能正常工作呢?
So why doesn't it just work?
Invoke() 产生 T.如果 GetValue 为 null，则 ? 运算符将 T 类型的返回值设置为 null，它不能.例如，如果 T 是 int，它不能使其为空 (int?)，因为需要的实际类型 (T = int) 不是.
Invoke() yields T. If GetValue is null, the ? operator sets the return value of type T to null, which it can't. If T is int for example, it can't make it nullable (int?) since the actual type required (T = int) isn't.
如果你把代码中的T改为int，你就会很清楚的看到问题所在.你问的最终结果是这样的:
If you change T to be int in your code, you will see the problem very clearly. The end result of what you ask is this:
get
{
int? x = GetValue?.Invoke();
return x.GetValueOrDefault(0);
}
这不是空传播运算符将为您做的事情.如果您恢复使用 default(T) 它确实知道该做什么，并且您避免了有问题的"空传播.
This is not something the null-propagation operator will do for you. If you revert to the use of default(T) it does know exactly what to do and you avoid the 'problematic' null-propagation.
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