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Published 20 hours ago • verified publisher icon zargony

Three parameters break closure highlighting

Open nathansizemore opened this issue 9 years ago • 4 comments

screen shot 2015-08-18 at 3 51 33 pm

nathansizemore avatar Aug 18 '15 19:08 nathansizemore

This doesn't look wrong to me? (Socket is not one of the highlighted std types)

zargony avatar Sep 14 '15 18:09 zargony

It's not the types, but the actual pipe '|'. The left one gets highlighted, but the right one does not. If there are less than 3 params, both get highlighted.

nathansizemore avatar Sep 14 '15 19:09 nathansizemore

Oops - missed that, thanks. (the left pipe symbol is mistakenly classified as an operator)

zargony avatar Sep 14 '15 20:09 zargony

With fd5447b, both pipes symbols are now classified equally. However they are still mistakenly seen as operators, which is because the rules do not include any closure syntax yet.

zargony avatar Sep 14 '15 23:09 zargony