bash - reuse code snippet in different function definitions

Question

Some of my utils functions often share the exact same initial code, that allows the function to be called with an argument or piped input:

RED=$(tput setaf 1)
YELLOW=$(tput setaf 3)
RESET=$(tput sgr0)

# -----------

function yellow {
  local arg
  if (( "$#" == 0 )); then
      IFS= read -r arg
      set -- "$arg"
  fi
  echo "${YELLOW}$1${RESET}"
}
function red {
  local arg
  if (( "$#" == 0 )); then
      IFS= read -r arg
      set -- "$arg"
  fi
  echo "${RED}$1${RESET}"
}

Is there a technique that allows me to "reuse" or somehow source the identical portion of code into the function definition? In my example this would be:

local arg
if (( "$#" == 0 )); then
  IFS= read -r arg     
  set -- "$arg"
fi

source script_name.sh or just . script_name.sh but be careful, script will be executed while sourcing — Ivan
– Ivan, Commented Feb 9, 2021 at 11:56
wow, did not know you could source random snippet of code. thanks, it works. Now I have to wrap my head around the possible side effects of the fact that the script is executed while sourcing. — Ghiro
– Ghiro, Commented Feb 9, 2021 at 12:14

Léa Gris · Accepted Answer · 2021-02-09 12:26:04Z

2

No, but you can make this much shorter with:

set -- "${1:-$(cat)}"

answered Feb 9, 2021 at 12:26

Léa Gris

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Comments

Sunderam Dubey · Accepted Answer · 2022-06-17 01:03:57Z

1

To partial source vars and/or functions from some script use this method:

$ cat test.sh
#!/bin/bash

test=123
mess() { echo ok; }
# Vars and functions ends here ^^
[[ $1 =~ source ]] && return
mess

Then souce it like this:

. ./test source

edited Jun 17, 2022 at 1:03

Sunderam Dubey

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answered Feb 9, 2021 at 12:33

Ivan

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