Shifting a block of text

user_17485413 0
Viewed 40

Trying to adapt the following function to use all arguments rather than just the first one.

The code should introduce nc spaces ahead of the start of each argument passed.

Each shifted argument must be on new line. The first argument is numeric, defining the amount of shift.

#!/bin/bash

shifted-block ()
{
 if (( $# == 1 )); then
   nc=0 ; arg="$1"
 elif (( $# >= 2 )); then
   nc="$1" ; arg="$2"
 else
   nc=0 ; arg="$1"
 fi

 nw=$(( nc + ${#arg} ))
 printf "%${nw}s\n" "$arg"
}
printf bash
2 Answers

Here is a solution :

shifted-block ()
{
    local nc=0
    if (($# > 1)); then nc=$1 ; shift ; fi

    local arg nw
    for arg in "$@" ; do
        nw=$(( nc + ${#arg} ))
        printf "%${nw}s\n" "$arg"
    done
}
user_17888510

Below is a version without using an explicit loop

shifted-block () {
    (( $# > 1 ))                            &&
    printf -v leading_blanks '%*s' "$1" ""  &&
    shift                                   &&
    printf '%s\n' "${@/#/$leading_blanks}"
}
user_138090010
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