prototype (::) syntax is ambigous

[bcherny]

i'm trying to check if an object has a prototype-

if a::b then c      # specific property check - ok
if a:: and b then c # no property specified - not ok

the latter example compiles to if (a.prototype.and(b)), but the intention was if (a.prototype && b).

the fix is to treat spaces around .prototype. vs. :: differently. ie.

a::b       #=> a.prototype.b
a:: b      #=> a.prototype(b)
a::b and c #=> a.prototype.b && c
a:: and c  #=> a.prototype && c

in the meantime, the workaround is to use parens-

if (a::) and b then c

[michaelficarra]

Duplicate of #1601. The unary form of :: will be removed... some day. Don't use it.

[bcherny]

sgtm. thanks for the quick response.