对“var”的质疑关键字和三元运算符 ?:

发布于 2024-11-16 16:54:06 字数 539 浏览 3 评论 0原文

如果 var 关键字在编译时解析，下面的代码如何工作？

class A {
}
class B : A {
}

int k = 1;
var x = (k < 0) ? new B() : new A();

编辑：
我终于明白问题不在于 var 本身，而在于 ?: 运算符的行为。出于某种原因，我认为以下情况是可能的：

object x = some ? 1 : ""

这根本不可能:)

相关问题（关于三元运算符）：
为什么在三元运算符中分配 null失败：null 和 int 之间没有隐式转换？

原文

If var keyword is resolved at compile time, how does the following work?

class A {
}
class B : A {
}

int k = 1;
var x = (k < 0) ? new B() : new A();

Edit:
I finally understood that the problem is not about the var itself, but about the behaviour of the ?: operator. For some reason, I thought that the following could be possible:

object x = something ? 1 : ""

and that's not possible at all :)

Related question (about ternary operator):
Why assigning null in ternary operator fails: no implicit conversion between null and int?

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梦屿孤独相伴 2024-11-23 16:54:06

结果是 A 类型，因为两个变量都是 A 类型，并且至少其中一个直接是 类型>A（不通过某种转换）。

编译器会检查三元表达式的两个部分，如果其中一个是另一个的子类型，则整个表达式将成为更通用的超类型。

但是，如果两者都不是直接的常见类型，则会发生编译器错误，可能是因为它不知道要为您向上转换多少（并且感觉不想找出）。

请参阅此处：

条件运算符 (?:) 根据布尔表达式的值返回两个值之一。以下是条件运算符的语法。
<前><代码>条件？第一个表达式：第二个表达式；
[...]
first_expression 和 secondary_expression 的类型必须相同，或者必须存在从一种类型到另一种类型的隐式转换。

The result is of type A, because both of the variables are of type A, and at least one of them is directly of type A (not through some conversion).

The compiler takes a look at both parts of the ternary expression, and if one of them is a subtype of the other, the entire expression becomes the more general supertype.

However, if neither is directly of the common type, then a compiler error occurs, probably because it doesn't know how much to upcast for you (and it doesn't feel like finding out).

See here:

The conditional operator (?:) returns one of two values depending on the value of a Boolean expression. Following is the syntax for the conditional operator.
condition ? first_expression : second_expression;
[...]
Either the type of first_expression and second_expression must be the same, or an implicit conversion must exist from one type to the other.

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