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Search Terms:
names of types still parse keywords
type name interprets reserved words
declaring names cannot use keywords
reserved words cannot be used
Expected behavior:
a member would be declared with the name of 'class', and I could access that member from other objects Actual behavior:
errors occur that prohibit the name being declared, however objects that attempt to access the member simply claim that it does not have a member of that name exported. Playground Link: Playground Related Issues:
None found
The text was updated successfully, but these errors were encountered:
I'm converting some JS code to TS. The JS module I'm working on exports a function with a "null" attribute added. The obvious export const null = ... produces a syntax error, and I've ended up here trying to find a workaround.
The solution @DanielRosenwasser gave may have worked in an earlier version of TypeScript, but in latest (3.2.4) I get a syntax error with this example code:
This issue has been marked as 'Question' and has seen no recent activity. It has been automatically closed for house-keeping purposes. If you're still waiting on a response, questions are usually better suited to stackoverflow.
TypeScript Version: 3.1.0-dev.20180907
Search Terms:
names of types still parse keywords
type name interprets reserved words
declaring names cannot use keywords
reserved words cannot be used
Code
Expected behavior:
a member would be declared with the name of 'class', and I could access that member from other objects
Actual behavior:
errors occur that prohibit the name being declared, however objects that attempt to access the member simply claim that it does not have a member of that name exported.
Playground Link:
Playground
Related Issues:
None found
The text was updated successfully, but these errors were encountered: