CODEWITHSUNDEEP
bash
Francesco Frassinelli
Francesco Frassinelli

Reputation: 3375

Bash should not count color codes as visible characters

I'm using fold to wrap my input file. I noticed that some colored lines are shorter. I found that bash counts color codes as characters, even if there are not visible. Example:

$ text="\e[0;31mhello\e[0m"; echo -e "$text - lenght ${#text}"
hello - lenght 18
$ text="hello"; echo -e "$text - lenght ${#text}"
hello - lenght 5

This happens for other non visible characters also:

$ text="a\bb\bc"; echo -e "$text - lenght ${#text}"
c - lenght 7

Is it possible to change this behavior? I would like that coreutils programs (bash and fold for example) could count only the visible characters.

Upvotes: 4

Views: 1727

Answers (2)

ton
ton

Reputation: 4597

Its not perfect, but you can remove the format bytes with a tool like sed before counting.

text="\e[0;31mhello\e[0m";
echo -e "# $text - lenght ${#text}";
# hello - lenght 18
x=$(echo -e "$text" | sed "s/$(echo -e "\e")[^m]*m//g");
echo "# $x - ${#x}"
# hello - 5

Upvotes: 1

Karoly Horvath
Karoly Horvath

Reputation: 96266

This is not a complete solution for your problem, but it's important to know that bash doesn't handle escape sequences in literals.

So "\b" is really 2 characters, \ and b. Only when you echo -e, then they are substituted.

Example:

text="a\bb\bc"
t=$(echo -e $text)
echo ${#t}
5  # the correct length

Upvotes: 4

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