Treat 'yield;' as 'yield undefined;' #22297
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ghost
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ghost
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mhegazy
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Mar 2, 2018
| ==== tests/cases/conformance/es6/yieldExpressions/generatorTypeCheck48.ts (1 errors) ==== | ||
| function* g() { | ||
| ~ | ||
| !!! error TS7025: Generator implicitly has type 'IterableIterator<any>' because it does not yield any values. Consider supplying a return type. |
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this was the correct error. without noImplicitAny, it should be an implicit any and not undefined.
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sandersn
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Mar 8, 2018
src/compiler/checker.ts
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| return aggregatedTypes; | ||
| } | ||
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| function getYieldedTypeOfYieldExpression(node: YieldExpression, isAsync: boolean, checkMode?: CheckMode): Type { | ||
| const errorNode = node.expression || node; | ||
| const expressionType = node.expression ? checkExpressionCached(node.expression, checkMode) : undefinedWideningType; |
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I appreciate the refactoring that is part of this PR -- it make the code much easier to read. But I want to point out that this is the only line needed to fix the bug. The bug would get fixed a lot faster if we could just review a one-line fix and look at the refactor separately.
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weswigham
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Mar 8, 2018
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Fixes #22290
Previously we would skip over
yield;with no argument; now we will treat it as yieldingundefined.