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Looks like `vararg` requires you to explicitly nam...
# announcements
c
cedric
05/27/2017, 5:21 AM
Looks like
vararg
requires you to explicitly name the following parameters. Sounds like a compiler weakness since inference is unambiguous here
➕ 1
i
ilya.gorbunov
05/27/2017, 12:09 PM
cedric: That's how vararg is designed to work.
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