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No. What I'm saying is, if type aliases do not cre...
# multiplatform
e
eddie
10/01/2017, 7:49 PM
No. What I'm saying is, if type aliases do not create new types, that means
actual typealias Foo = Bar
is equivalent to
typealias Foo = Bar
. In which case, why can't
fun fooBar()
fulfill what would require
actual fun fooBar()
?
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