Open nathansizemore opened 9 years ago
This doesn't look wrong to me? (Socket is not one of the highlighted std types)
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.
Oops - missed that, thanks. (the left pipe symbol is mistakenly classified as an operator)
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.