bash printing and incrementing array values

Question

I'm making a bash script in which I need to print a number while it's incremented like this:

I have made this but is not working:

#!/bin/bash
i=0
pass[0]=0
pass[1]=0
pass[2]=0
pass[3]=0
for i in $(seq 1 9)
    pass[3]="$i"
    echo ${pass[*]}
done

I paste the script on cli and i get this.

$ ~ #!/bin/bash
$ ~ i=0
$ ~ pass[0]=0
$ ~ pass[1]=0
$ ~ pass[2]=0
$ ~ pass[3]=0
$ ~ for i in $(seq 1 9)
>     pass[3]="$i"
bash: error sintáctico cerca del elemento inesperado `pass[3]="$i"'
$ ~     echo ${pass[*]}
0 0 0 0
$ ~ done
bash: error sintáctico cerca del elemento inesperado `done'
$ ~

Do the values have to be in an array or can you just generate them as needed? — Ray Toal
– Ray Toal, Commented Nov 26, 2012 at 17:16
Your for i in ... is missing the corresponding do - e.g. for i in $(seq 1 9); do pass[3]="$i"; done — twalberg
– twalberg, Commented Nov 26, 2012 at 17:25
there are many ways to accomplish this task, but to make the least changes and answer what is wrong with your code, you forgot 'do' after the for expression. — matchew
– matchew, Commented Nov 26, 2012 at 17:26

anubhava · Accepted Answer · 2012-11-26 17:13:18Z

6

Use this pure bash script:

for ((i=0; i<10; i++)); do
   printf "%04d\n" $i
one

OUTPUT:

answered Nov 26, 2012 at 17:13

anubhava

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3 Comments

Ray Toal Over a year ago

+1 but one can also do for i in {1..10}; do printf "%04d\n" $i; done. Not sure which style of for-loop is better; I just like expansion.

anubhava Over a year ago

@RayToal: I just prefer for ((i=0; i<10; i++)); because it also accepts variable like N=10; for ((i=0; i<N; i++)); as well.

Ray Toal Over a year ago

Yes definitely more flexible in general.

matchew · Accepted Answer · 2012-11-26 17:13:35Z

2

#!/bin/bash
i=0
pass[0]=0
pass[1]=0
pass[2]=0
pass[3]=0
for i in $(seq 1 9)
do
    pass[3]="$i"
    echo ${pass[*]}
done

did you forget 'do'

answered Nov 26, 2012 at 17:13

matchew

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2 Comments

gniourf_gniourf Over a year ago

That's a very inefficient way of doing it. Prefer anubhava's answer.

matchew Over a year ago

@gniourf_gniourf as do I. but I was addressing why his sample code was not working. He had excluded the do

gniourf_gniourf · Accepted Answer · 2012-11-26 17:55:41Z

2

For those of you who like expansions, you can also do:

printf "%s\n" {0001..0009}

or

printf "%.4d\n" {1..9}

No loop!

You can store in an array thus:

$ myarray=( {0001..0009} )
$ printf "%s\n" "${myarray[@]}"
0001
0002
0003
0004
0005
0006
0007
0008
0009
$ echo "${myarray[3]}"
0004

answered Nov 26, 2012 at 17:55

gniourf_gniourf

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Comments

rici · Accepted Answer · 2012-11-27 06:12:03Z

1

You can do the formatting with seq:

seq -w 0000 0010

(if you don't like the {0000..0010} notation, which is more efficient but doesn't allow parameter substitution.)

answered Nov 27, 2012 at 6:12

rici

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