Ok, given that you know perl knows that function is variadic, you know it's going to try to put everything following it into its parameter list. My understanding is that the parser is only able to look ahead one token at a time. The fact that it got two tokens ahead before it found out it couldn't parse things means it just couldn't continue. It isn't capable of backing up from function( / 4; to function() / 4;. You'll just have to provide that information to perl soon enough for the parser to use it. Sorry. Either use the parameter list or set the prototype.

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In reply to Re^3: Unexpected parser error by diotalevi
in thread Unexpected parser error by saintmike

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