为什么可空布尔变量不允许的话（可为空），但也允许的话（可为空==真）？

问题描述

这code编译：

    static void Main(string[] args)
    {
        bool? fred = true;

        if (fred == true)
        {
            Console.WriteLine("fred is true");
        }
        else if (fred == false)
        {
            Console.WriteLine("fred is false");
        }
        else
        {
            Console.WriteLine("fred is null");
        }
    }

这code做的没有的编译。

This code does not compile.

    static void Main(string[] args)
    {
        bool? fred = true;

        if (fred)
        {
            Console.WriteLine("fred is true");
        }
        else if (!fred)
        {
            Console.WriteLine("fred is false");
        }
        else
        {
            Console.WriteLine("fred is null");
        }
    }

我想如果（booleanEx pression == true）而被认为是一个冗余。为什么不是在这种情况下？

I thought if(booleanExpression == true) was supposed to be a redundancy. Why isn't it in this case?

推荐答案

有没有从隐式转换可空＆LT;布尔＆GT; 到布尔。这里的是的距离布尔的隐式转换为可空＆LT;布尔＆GT; ，这就是发生了什么（在语言方面）的每个在第一个版本的布尔常量。该布尔运算符==（可空＆LT;布尔＆gt;中可空＆LT;布尔＆GT; 运营商，然后应用（这是不太一样的对方解除运营商 - 结果却布尔，不是可空＆LT;布尔＆GT; ）

There's no implicit conversion from Nullable<bool> to bool. There is an implicit conversion from bool to Nullable<bool> and that's what happens (in language terms) to each of the bool constants in the first version. The bool operator==(Nullable<bool>, Nullable<bool> operator is then applied. (This isn't quite the same as other lifted operators - the result is just bool, not Nullable<bool>.)

在换句话说，EX pression'弗雷德==假的类型是布尔，而EX pression弗来德是类型可空＆LT;布尔＆GT; 因此你不能用它作为如果EX pression

In other words, the expression 'fred == false' is of type bool, whereas the expression 'fred' is of type Nullable<bool> hence you can't use it as the "if" expression.

编辑：要回答的意见，从未有从可空℃的隐式转换; T＆GT; 到 T 和很好的理由 - 转换隐不应该抛出异常，除非你想空隐式转换为默认（T）有没有很多东西可以做。

To answer the comments, there's never an implicit conversion from Nullable<T> to T and for good reason - implicit conversions shouldn't throw exceptions, and unless you want null to be implicitly converted to default(T) there's not a lot else that could be done.

此外，如果有的是的隐式转换左右逢源轮，一个前pression像可空+非空将是非常混乱的（对于支持+，像 INT ）。既+（ΔT1，T）+和（T，T）将可用，这取决于操作数被转换！ - 但结果可能非常不同

Also, if there were implicit conversions both ways round, an expression like "nullable + nonNullable" would be very confusing (for types that support +, like int). Both +(T?, T?) and +(T, T) would be available, depending on which operand were converted - but the results could be very different!

我的决定，只有从显式转换可空＆LT后面100％; T＆GT; 到 T

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