I'm making a conditional type using the ?operator in TypeScript. Based on a condition I want the type to be two different types. When doing this, I found, that the condition affects the branches.

我正在使用TypeScrip中的？运算符创建条件类型。根据一个条件，我希望类型是两种不同的类型。在这样做的时候，我发现这种情况会影响到树枝。

The MRE is here:

MRE在这里：

type Type1<O> = O extends never ? unknown : (undefined extends O ? 1 : 2);
type a = Type1<string | undefined>

type Type2<O> = undefined extends O ? 1 : 2;
type b = Type2<string | undefined>

or in TS Playground.

或者在TS游乐场。

I would expect a and b to be the same, as the condition O extends never should never be true, unless O is never. This is however not the case as type a = 1 | 2 and type b = 1;

我希望a和b是相同的，因为O扩展的条件永远不应该为真，除非O永远不为真。然而，当类型a=1|2和类型b=1时，情况并非如此；

I cannot think of a reason, why the type union would be present, unless the condition in Type1 somehow iterates over the string | undefined uninon.

我想不出为什么会出现类型联合，除非Type1中的条件以某种方式迭代了字符串|unfined uninon。

Why is this the case? How can I work around this, if I want a to be only 1?

为何会是这样呢？如果我希望a仅为1，我该如何解决这个问题？

You are being surprised by distribution over union. Because you are using a naked type in left-hand side position in a conditional type, what you wrote is basically equivalent to writing Type1<string> | Type1<undefined>. All you need to do is wrap the operands in square brackets.

你对分配而不是工会感到惊讶。因为您在条件类型中的左侧位置使用的是裸类型，所以您编写的内容基本上等同于编写Type1<字符串>|Type1 。您所需要做的就是将操作数括在方括号中。