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Parse backtick delineated string without interpolation as string literals #18109
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- changed the title
[-]Parse backticks correctly when using for all strings (for consistency)[/-][+]Parse backtick delineated string without interpolation as string literals[/+]on Aug 29, 2017 I'd argue that this is a duplicate of #16592 and that that particular issue wasn't fully fixed by #17704.
Daniel Rosenwasser (@DanielRosenwasser)
We should just allow people to use NoSubstitutionTemplates the way that they're allowed to use regular strings.
That also includes being able to put
NoSubstitutionTemplates as types, and anywhere else that string literals can be used.closing in favor of #16592. thanks.
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on Jun 14, 2018
TypeScript Version: 2.4.0
Code
Expected behavior:
testwill be a variable of type"a" | "b", like when single quotes or double quotes are used.Actual behavior:
It does not compile, giving various errors.
Justification:
We like to enforce the use of template strings instead of concatenation, but that leaves us with a mixture of backtick strings and either double or single quote strings. Ideally we would enforce just one type of quote throughout our projects and leave it at that. In order to utilize the power of template strings, we should obviously enforce only backticks for strings.
The problem now is that the compiler doesn't accurately parse strings delineated by backticks. When backtick delineated strings don't have any interpolation in them, the compiler should be able to recognize that they are just literal strings and should use them for type inference and everything else.