Can't index of the "#[hello]" in bash

Question

The following index_of function doesn't work for all cases:

#!/bin/bash

index_of() {
    local string="$1"
    local search_string="$2"
    
    local prefix=${string/${search_string}*/}
    
    local index=${#prefix}
     
    if [[ index -eq ${#string} ]];
    then
        index=-1
    fi
    
    printf "%s" "$index"
}

a='#[hello] world'

b=$(index_of "$a" "world")
echo "A: $b"

b=$(index_of "$a" "hello")
echo "B: $b"

b=$(index_of "$a" "#[hello]")
echo "C: $b"

Here is the output:

A: 9
B: 2
C: -1

The A and B is correct, but the C is incorrect.

The C should be 0 instead of -1 .

What's wrong in the index_of function and how to fix the C index?

Answer 1

#!/bin/bash

index_of() {
    local string="$1"
    local search_string="$2"

    local prefix=${string/${search_string}*/}

    local index=${#prefix}

    if [[ $index -eq ${#string} ]];
    then
        index=-1
    fi

    printf "%s" "$index"
}

a='#[hello] world'

b=$(index_of "$a" "world")
echo "A: $b"

b=$(index_of "$a" "hello")
echo "B: $b"

b=$(index_of "$a" "\#\[hello\]")
echo "C: $b"

result

A: 9
B: 2
C: 0

Answer 2

When using parameter expansion in the pattern part of constructs like ${param/pattern/repl} , quote the parameter expansion to remove the special meaning of any shell pattern metacharacters that may exist in the parameter.

Here's a slightly different implementation of your index_of function. In case the length of the second argument is zero, the index shall be zero. At the end, a return value is provided to indicate success or failure. "$2" is quoted in ${1/"$2"*/} so it is treated literally, not as a shell pattern.

#! /bin/bash -

index_of () {
    local idx pfx

    pfx=${1/"$2"*/}
    idx=$((${#1} == ${#pfx} ? -!!${#2} : ${#pfx}))

    printf '%s\n' "$idx"
    return "$((idx < 0))"
}

for str in world hello '#[hello]'; do
    index_of '#[hello] world' "$str"
done

output:

9
2
0