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hildjj avatar hildjj commented on September 15, 2024

As it turns out, the type there is correct. See this test as an example. In this case, the output of generate changes based on the value of output. If I had it to do over again or didn't care about backward-compatibility, I think I would have made different method names that returned different types. As-is, the types are more complex to understand than is ideal.

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siefkenj avatar siefkenj commented on September 15, 2024

I am confused. Is there somewhere else in the types file that allows output to be something other than "parser"? In the documentation line directly above line 1193 it says that several strings besides "parser" are permitted. TS also complained in my project when I passed output: "source" (I had to use a // @ts-ignore to suppress the error...)

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hildjj avatar hildjj commented on September 15, 2024

Yes. There are several overloads for generate, selected based on the different types of the options parameter. Each of those return something different based on what output you select.

export function generate(grammar: string, options?: ParserBuildOptions): Parser;
export function generate(grammar: string, options: SourceBuildOptions<"source">): string;
export function generate(grammar: string, options: SourceBuildOptions<"source-with-inline-map">): string;
export function generate(grammar: string, options: SourceBuildOptions<"source-and-map">): SourceNode;
export function generate(grammar: string, options: SourceBuildOptions<SourceOutputs>): SourceNode | string;
export function generate(grammar: string, options: SourceOptionsBase<"ast">): ast.Grammar;

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