Greetings, I present an elementary question for a bored monk.
$ perl -e 'print "\@hi\n";' @hi

I understand why the backslash before the @ is required here to get a literal @ in an interpolating string: without it, the @ would be interpreted as indicating an array named hi.

It is not needed (though adding it is harmless) in strings where there is no ambiguity about what the @ might represent:

$ perl -e 'print "@=hi\n";' @=hi
But why is it needed to signal a literal @ in the following?
$ perl -e 'print "@$0\n";' $ perl -e 'print "\@$0\n";' @-e

How can an @ followed by a $ be interpreted as a valid array name? (The reverse, $@, is a predefined variable, but @$ is not.) And if it can't, why is the backslash needed to indicate this @ is literal?


In reply to escaping the @ in an interpolating string by raygun

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