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efficient way to use matched string in variable substitution
From: |
L A Walsh |
Subject: |
efficient way to use matched string in variable substitution |
Date: |
Mon, 23 Aug 2021 11:36:52 -0700 |
User-agent: |
Thunderbird 2.0.0.24 (Windows/20100228) |
Starting with a number N, is there
an easy way to print its digits into an array?
I came up with a few ways, but thought this
would be nice (with '\1' or '$1' being what was matched
in the 1st part), this could be statement:
arr=(${N//[0-9]/\1 })
or
arr=(${N//[0-9]/$1 })
Instead of using loops (my=declare):
n=988421
for x in 0 1 2 3 4 5 6 7 8 9;do n=${n//$x/$x }; done
arr=($n)
my -p arr
declare -a arr=([0]="9" [1]="8" [2]="8" [3]="4" [4]="2" [5]="1")
or w/substrings:
for ((d=0; d<${#n};d+=1)); do arr+=(${n:$d:1}); done
my -p arr
declare -a arr=([0]="9" [1]="8" [2]="8" [3]="4" [4]="2" [5]="1")
Not a big thing, but having some way for the match of an RE
to be specified in the output would be handy...
- efficient way to use matched string in variable substitution,
L A Walsh <=
- Re: efficient way to use matched string in variable substitution, Greg Wooledge, 2021/08/23
- Re: efficient way to use matched string in variable substitution, L A Walsh, 2021/08/23
- Re: efficient way to use matched string in variable substitution, Greg Wooledge, 2021/08/23
- Re: efficient way to use matched string in variable substitution, Léa Gris, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Greg Wooledge, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Léa Gris, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Mike Jonkmans, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Greg Wooledge, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Mike Jonkmans, 2021/08/24
- Re: efficient way to use matched string in variable substitution, Léa Gris, 2021/08/24