I simply can't figure out how does this line of awk code work.
$$1~/^[^0-9]/ {print $$1;}
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I wouldn't be super surprised if this were from a makefile, though. – Benjamin W. Sep 19 '19 at 14:34
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The following might be relevant: [Why does AWK not work correctly in a makefile?](https://stackoverflow.com/q/30445218/8344060) – kvantour Sep 19 '19 at 14:44
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The $ sign is an operator in awk to reference a field. Example:
$ echo "foo bar car" | awk '{print $2}'
bar
This prints bar, as bar is the content of the second field.
A double dollar sign is actually a double reference that will use the information of the first field reference to get to the other field reference. Example:
$ echo "foo bar car 1 2 3" | awk '{print $$5}'
bar
$ echo "foo bar car 1 2 3" | awk '{print $5}'
2
Here it prints bar as $5 is dereferenced as 2 and thus $$5 is equivalent to $2
kvantour
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3Ta make it simpler to read/understand, you could use parentheses like this: `awk '{print $($5)}'` It will expand the parentheses first, giving `2`, and then you get `$2` that gives `bar`. `echo "foo 3 car 1 2 3" | awk '{print $$$5}' => car` or `echo "foo 3 car 1 2 3" | awk '{print $($($5))}' = car` – Jotne Sep 19 '19 at 15:36