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I am new at unix, I'm trying to learn the bash language, and when I went to the "Testing expression", I found this one:

[[ "whatever" =~ h[aeiou] ]]

I already read the answer to this question, so I understand what the operator =~ does. After running the previous command, the output of echo $? is 0, meaning the condition inside the [[ ]] is met. If instead I type

[[ "whatever" =~ h[sdfghjkl] ]]

the output of echo $? is 1, so the condition was not met.

So, I would like to know what the h[aeiou] and h[sdfghjkl] are. Is h a predefined function inside the [[ ]] expression? And if so, what is it doing? If not, what h[aeiou] and h{sdfghjkl] actually are?

Thanks.

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  • Your first command will evaluate as true if the string "whatever" contains the letter h followed by a vowel. Second command looks for h followed by your list of letters. h is not a predefined function but is part of the regular expression to match against.
    – doneal24
    Commented Oct 11, 2022 at 20:33
  • They are regular expressions. h[aeiou] matches a pair of letters, the left letter being h and the right-hand letter being one of those listed between [ and ] (an English vowel letter). h[sdfghjkl] is exactly the same, but with matching the right-hand letter with a list of different letters between [ and ].
    – Sotto Voce
    Commented Oct 11, 2022 at 20:33

1 Answer 1

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Like it says in the other post, the =~ does a pattern match, with the right-side operand being a regular expression (regex).

The brackets in h[aeiou] are part of the regex syntax, there's no named function there. h[aeiou], matches an h and then any single one of aeiou, so whatever matches because it contains an h followed by an a.

You're going to hear about regexes, so it might be worth looking into them a bit:

(The Wikipedia page seems to also go in the formal details, don't get too caught up in that.)

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  • Thanks for the quick answer! It solved my question, and thanks for the provided links. :]
    – andr.lex
    Commented Oct 11, 2022 at 20:58
  • @andr.lex, note that =~ doesn't require a full-string match, so indeed [[ whatever =~ h[aeiou] ]] matches. That's unlike most contexts where shell patterns are used. (The original version of this answer, before terdon fixed it, said it wouldn't match, and I don't know what I was thinking there.)
    – ilkkachu
    Commented Oct 12, 2022 at 17:27

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