Neither does @foo get converted to a scalar, nor is + defined differently for @foo than it is for $bar.

The value of @foo is a number in scalar context, but it's not a conversion - @foo doesn't change. Nor is + internally a huge dispatch table, doing all kinds of different things depending on the types of its arguments. Binary + always gets two scalars as arguments, and isn't doing anything before the scalar values are known.

In reply to Re^2: Static typing is mostly a waste of time by Anonymous Monk
in thread Static typing is mostly a waste of time by johnnywang

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