Syntax with pound and percent sign after shell parameter name

Question

Please explain the following syntax:

  i##*.

  i%.*

I understand what it's doing, but I want to know the general pattern (why/how is it doing so).

Code where it appears:

#!/bin/bash

recursive_name_change()
{
    cd "$1"
    for i in *
    do
        #echo "${i##*.}"
        if [ -d "$i" ]
        then
            recursive_name_change "$i"
        elif [ "${i##*.}" = "cpp" ]
        then
             new_name=${i%.*}".c"
        mv "$i" "$new_name"
        fi
        done
        cd ../
    }

recursive_name_change .

Someone please also suggest, where can I find these peculiar syntax forms from?

Did you even *try* to read the bash man page and search for `##`? — ghoti, Mar 16 '16 at 13:27
No, I didn't search it in the bash manual, but I did perform google search but did not have any useful results. — Skynet094, Mar 16 '16 at 13:38
@Dunatotatos (suggested edit) This is *not* exclusively Bash, see the POSIX shell specification. — Benjamin W., Mar 16 '16 at 13:40
@BenjaminW. Oh yes, you are right. My mistake. I saw that in the bash man, and automatically thought it was for bash only. — Dunatotatos, Mar 16 '16 at 16:00

choroba · Accepted Answer · 2016-03-16T13:17:56.687

See Parameter Expansion in man bash:

   ${parameter#word}
   ${parameter##word}
Remove matching prefix pattern. The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches the beginning of the value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the # case) or the longest matching pattern (the ## case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.
  ${parameter%word}
  ${parameter%%word}
Remove matching suffix pattern. The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the % case) or the longest matching pattern (the %% case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each mem- ber of the array in turn, and the expansion is the resultant list.

In short, # removes the pattern from the left, % from the right, doubling the symbol makes the matching greedy. (Mnemonic: # is to the left of % on most keyboards).

@ceving, not just English keyboards, also keyboards found in Latin America and Spain. Google image search indicates also keyboards in India and China, so "most keyboards" is probably accurate. — Josh, May 04 '20 at 17:07

score 1 · Answer 2 · edited Mar 16 '16 at 16:05

1

Look for the section "Substring removal" in this manual:

${string##substring}
Deletes longest match of $substring from front of $string.
${string%substring}
Deletes shortest match of $substring from back of $string.

edited Mar 16 '16 at 16:05

Benjamin W.

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answered Mar 16 '16 at 13:13

cd1

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The ABS is generally *not* a good resource. It ignores many caveats. Ignores best practices. Includes poor quality code. etc. – Etan Reisner Mar 16 '16 at 14:21
Yes, but does so by using `echo ${unquoted_variable}` which is a poor practice to show. – Etan Reisner Mar 16 '16 at 18:07

Syntax with pound and percent sign after shell parameter name

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