英文:
Is type inference broken when a type guard is used in a if statement
问题
这段代码存在类型推断问题，还是我漏掉了什么？
将这段代码：
export type Writable<T> = {-readonly [K in keyof T]: T[K];};class Foo {readonly foo = 'foo';}class Bar {readonly bar = 'bar';}type WritableFooBar = Writable<Foo | Bar>;declare function isFoo(obj: any): obj is Foo;function test(foobar: WritableFooBar): void {if (!isFoo(foobar)) {} // 也会出现在if(isFoo())中foobar// ^? Foo | Writable<Bar>}
翻译为：
这段代码存在类型推断问题，还是我漏掉了什么？
英文:
Take this code :
export type Writable<T> = {-readonly [K in keyof T]: T[K];};class Foo {readonly foo = 'foo';}class Bar {readonly bar = 'bar';}type WritableFooBar = Writable<Foo | Bar>;declare function isFoo(obj: any): obj is Foo;function test(foobar: WritableFooBar): void {if (!isFoo(foobar)) {} // happens also with if(isFoo())foobar// ^? Foo | Writable<Bar>}
Is type inference broken in this case or am I missing something ?
答案1
得分: 0
这实际上是一个在4.8版本中出现的回归问题，在5.0版本中已修复。
英文:
This actually a regression that appeared in 4.8 and was fixed in 5.0.
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