为什么 null 需要在这里进行显式类型转换？
以下代码无法编译：
//int a = ...
int? b = (int?) (a != 0 ? a : null);
为了编译，需要将其更改为
int? b = (a != 0 ? a : (int?) null);
由于 b = null 和 b = a 都是合法的，所以这没有意义大部头书。
为什么我们必须将 null 转换为 int？ 为什么我们不能简单地为整个表达式提供显式类型转换（我知道这在其他表达式中是可能的）例）？
The following code does not compile:
//int a = ...
int? b = (int?) (a != 0 ? a : null);
In order to compile, it needs to be changed to
int? b = (a != 0 ? a : (int?) null);
Since both b = null and b = a are legal, this doesn't make sense to me.
Why do we have to cast the null into an int? and why can't we simply provide an explicit type cast for the whole expression (which I know is possible in other cases)?
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来自 C# 语言规范第 7.13 章：
在您的情况下，没有从 int 到 null 的隐式转换，也没有相反的转换。你的强制转换解决了问题，int可以转换为int吗？
From chapter 7.13 of the C# Language Specification:
In your case, there is no implicit conversion from int to null nor the other way around. Your cast solves the problem, int is convertible to int?
?:运算符的两个替代项必须是同一类型。否则，编译器无法推断出整个条件表达式的类型。
null不是int，因此您需要向编译器提示结果类型是int？。编辑：正如其他人指出的那样，这两种类型不需要相同，但其中一种类型应该可以转换为另一种类型（另一种类型将是结果类型）。有关更多详细信息，请参阅规范。
The both alternatives of
?:operator must be of the same type.Otherwise, the compiler cannot deduce the type of the whole conditional expression.
nullis not anint, so you need to give a hint to the compiler that the resulting type isint?.Edit: as the others pointed out, the two types don't need to be the same, but one of them should be castable to another (that another one will be the result type). See specs for more details.
如果使用
default(int?)而不是null，则可以避免强制转换。You can save yourself from casting if you use
default(int?)instead ofnull.这是因为当你使用这种表示法时，两个成员必须是相同的类型，所以你必须明确地说“这个成员是一个 int？”。
清楚了吗？
it's because when you use that kind of notation, both member must be of the same type, so you have to explicitly say "this member is a int?".
is it clear ?
当您将条件运算符与不同类型的操作数一起使用时，编译器将检查其中一种类型是否可以隐式转换为另一种类型。
如果这两种类型都不能隐式转换为另一种类型，则会给出错误，即使存在它们都可以隐式转换为的第三种类型。编译器不会同时在两侧插入隐式转换。
在您的情况下，
int和null都需要隐式转换才能成为int?，因此它不起作用。您需要更改代码，以便一侧是
int?，另一侧可以隐式转换为int?。最简单的方法是将null替换为new int?()，如下所示：这只需要一次隐式转换（
int到int？）。When you use the conditional operator with operands of different types, the compiler will check whether one of the types can be implicitly converted to the other type.
If neither type can be implicitly converted to the other one, it will give an error, even if there is a third type that they can both implicitly convert to. The compiler will not insert an implicit conversion on both sides at once.
In your case, both
intandnullrequires implicit conversions to becomeint?, so it doesn't work.You need to change your code so that one side is an
int?and the other side can be implicitly converted toint?. The simplested way to do this is to replacenullwithnew int?(), like this:This only requires one implicit conversion (
inttoint?).这几乎是我关于条件运算符的一个问题的重复。它不是 null，它实际上是
:。那里接受的答案非常好，来自 C# 编译器团队的 Eric Lippert。
This is pretty much a duplicate of one of my questions, regarding the conditional operator. It's not the null, it really is the
:.The answer accepted there is pretty good, from none other than Eric Lippert, who is on the C# compiler team.