microsoft / microsoft/TypeScript
Assertion methods (`asserts this is`) are not CFA'd without error
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描述
TypeScript Version: 4.0.5, 4.1.0-beta, 4.2.0-dev.20201112
Search Terms:
- Multiple assertion methods
- "asserts this is"
- assertion "cfa"
Code
class X<T, Locked extends boolean> {
public x: null | T | (Locked extends true ? 1 : 2) = null
public assert<U>(): asserts this is X<U, Locked> { }
public lock(): asserts this is X<T, true> {}
}
const x: X<string | number, false> = new X<string | number, false>()
x.assert<string>()
// x is X<string, false> here
x.lock()
Expected behavior:
Either:
- After
x.lock(),xis narrowed toX<string, true> - Or, the
x.lock()line throws at compile-time due to not being CFA'd
Actual behavior:
After x.lock(), x is narrowed to X<string, false> & X<string | number, true>.
Analysis:
x being typed as X<string, false> & X<string | number, true> shows that the x.lock() narrows from the original type (X<string | number, false>) instead of the narrowed type at that position (X<string, false>).
If instead a "top-level" assertion function is used the type is properly narrowed:
class X<T, Locked extends boolean> {
public x: null | T | (Locked extends true ? 1 : 2) = null
public assert<U>(): asserts this is X<U, Locked> { }
public lock(): asserts this is X<T, true> {}
}
const x: X<string | number, false> = new X<string | number, false>()
x.assert<string>()
// x is X<string, false> here
declare function lock<T>(v: X<T, boolean>): asserts v is X<T, true>
lock(x)
This leads me to believe this is an issue with the x.lock() call not being CFA'd, in which case the correct behavior would be to throw ts(2775) on the x.lock() line.
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调研方向
首先运行提供的 TypeScript Playground 复现代码,并比较 x.assert<string>() 和 x.lock() 之后的控制流 narrowing。调查编译器对 asserts this is 方法以及所报告的交叉类型的处理方式。当第二个 assertion 要么 narrowing 为 X<string, true>,要么报告预期的编译时错误,并且该示例具有回归覆盖时,即视为完成。
由索引模型根据 Issue 内容生成。
评估
- 技术栈
- typescript
- 领域
- compilers
- Issue 类型
- 缺陷
- 难度
- 4/5
- 预计耗时
- 3-5 天
- 活跃度
- 停滞
- 描述清晰度
- 基本清楚
- 新手友好度
- 35/100