Closed eeasley closed 11 years ago
const FOO = \bar let baz = "$FOO" let qux = "$(FOO)"
transpiles to
__strnum = ... __typeof = ... baz = __strnum(FOO); qux = "bar";
That seems like unexpected and undesirable behavior to me.
Yep, this is a bug. Both baz and qux should evaluate as qux does now.
Thanks.
transpiles to
That seems like unexpected and undesirable behavior to me.