I guess this comes from the fact that do is also used for flow control, which means it doesn't fall under the "Looks like a function" rule. print "Hello" if(0) || 1; prints "Hello", because if is not a function.

This behaviour is actually documented for last and its kind:

It is also exempt from the looks-like-a-function rule, so last ("foo")."bar" will cause "bar" to be part of the argument to last.
Maybe the documentation for do should mention the same thing.

In reply to Re: parser bug handling 'do ($foo)->{bar}' ? by Eily
in thread parser bug handling 'do ($foo)->{bar}' ? by LanX

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