How to Use an array of strings in bash

up vote
1
down vote

favorite

I writte the following script

load function should set the array disk[a]=1 and disk[b]=2 and so on

Then the function out should print the array $disk[a] , and $disk[b] , and so on

But what we get from function out is always the number 4

Instead, I want to get the following:

What is wrong here ?

How to fix it so function out will print:

the script:

#!/bin/bash

function load

 counter=1
 for input in a b c d
 do
 disk[$input]=$counter
 let counter=$counter+1
 echo $disk[$input]
 done



function out

 counter=1
 for input in a b c d
 do
 echo $disk[$input]
 let counter=$counter+1
 done


echo "run function load"
load
echo "run function out"
out

the output:

./test
run function load
1
2
3
4
run function out
4
4
4
4

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

add a commentÂ |Â

up vote
1
down vote

favorite

I writte the following script

load function should set the array disk[a]=1 and disk[b]=2 and so on

Then the function out should print the array $disk[a] , and $disk[b] , and so on

But what we get from function out is always the number 4

Instead, I want to get the following:

What is wrong here ?

How to fix it so function out will print:

the script:

#!/bin/bash

function load

 counter=1
 for input in a b c d
 do
 disk[$input]=$counter
 let counter=$counter+1
 echo $disk[$input]
 done



function out

 counter=1
 for input in a b c d
 do
 echo $disk[$input]
 let counter=$counter+1
 done


echo "run function load"
load
echo "run function out"
out

the output:

./test
run function load
1
2
3
4
run function out
4
4
4
4

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

add a commentÂ |Â

up vote
1
down vote

favorite

I writte the following script

load function should set the array disk[a]=1 and disk[b]=2 and so on

Then the function out should print the array $disk[a] , and $disk[b] , and so on

But what we get from function out is always the number 4

Instead, I want to get the following:

What is wrong here ?

How to fix it so function out will print:

the script:

#!/bin/bash

function load

 counter=1
 for input in a b c d
 do
 disk[$input]=$counter
 let counter=$counter+1
 echo $disk[$input]
 done



function out

 counter=1
 for input in a b c d
 do
 echo $disk[$input]
 let counter=$counter+1
 done


echo "run function load"
load
echo "run function out"
out

the output:

./test
run function load
1
2
3
4
run function out
4
4
4
4

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

I writte the following script

load function should set the array disk[a]=1 and disk[b]=2 and so on

Then the function out should print the array $disk[a] , and $disk[b] , and so on

But what we get from function out is always the number 4

Instead, I want to get the following:

What is wrong here ?

How to fix it so function out will print:

the script:

#!/bin/bash

function load

 counter=1
 for input in a b c d
 do
 disk[$input]=$counter
 let counter=$counter+1
 echo $disk[$input]
 done



function out

 counter=1
 for input in a b c d
 do
 echo $disk[$input]
 let counter=$counter+1
 done


echo "run function load"
load
echo "run function out"
out

the output:

./test
run function load
1
2
3
4
run function out
4
4
4
4

linux bash shell-script array

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

edited 3 hours ago

Jeff Schaller

35.1k952115

edited 3 hours ago

Jeff Schaller

35.1k952115

edited 3 hours ago

Jeff Schaller

35.1k952115

asked 3 hours ago

yael

2,1401650

asked 3 hours ago

yael

2,1401650

asked 3 hours ago

yael

2,1401650

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

Bash arrays are indexed arrays by default:

An indexed array is created automatically if any variable is assigned to using the syntax name[subscript]=value

... but you are using letters as the index, so you probably want an associative array, which means you need an:

declare -A disk

before calling the functions.

answered 3 hours ago

Jeff Schaller

35.1k952115

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

add a commentÂ |Â

up vote
3
down vote

Look at what happens to the array when you initialize it:

$ i=0; for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -a A=([0]="3")

Only one element is present, and it has the index zero.

By default, arrays are indexed by numbers, and the numerical values of the indices you used happened to be all zero. Actually, in an arithmetic context, like the subscript of a regular array, a string is taken as the name of a variable, and the value of that variable is used. So, if we set a, b... to numbers, then we get something different:

$ a=123; b=456; c=789; d=999; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done;
$ declare -p A
declare -a A=([123]="0" [456]="1" [789]="2" [999]="3")

To actually use the strings themselves as indices, declare the array as an associative array first with declare -A arrayname or typeset -A arrayname:

$ unset A; declare -A A; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -A A=([a]="0" [b]="1" [c]="2" [d]="3" )

answered 3 hours ago

ilkkachu

53.3k780146

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

accepted

Bash arrays are indexed arrays by default:

An indexed array is created automatically if any variable is assigned to using the syntax name[subscript]=value

... but you are using letters as the index, so you probably want an associative array, which means you need an:

declare -A disk

before calling the functions.

answered 3 hours ago

Jeff Schaller

35.1k952115

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

add a commentÂ |Â

up vote
4
down vote

accepted

Bash arrays are indexed arrays by default:

An indexed array is created automatically if any variable is assigned to using the syntax name[subscript]=value

... but you are using letters as the index, so you probably want an associative array, which means you need an:

declare -A disk

before calling the functions.

answered 3 hours ago

Jeff Schaller

35.1k952115

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

add a commentÂ |Â

up vote
4
down vote

accepted

Bash arrays are indexed arrays by default:

An indexed array is created automatically if any variable is assigned to using the syntax name[subscript]=value

... but you are using letters as the index, so you probably want an associative array, which means you need an:

declare -A disk

before calling the functions.

answered 3 hours ago

Jeff Schaller

35.1k952115

Bash arrays are indexed arrays by default:

An indexed array is created automatically if any variable is assigned to using the syntax name[subscript]=value

... but you are using letters as the index, so you probably want an associative array, which means you need an:

declare -A disk

before calling the functions.

answered 3 hours ago

Jeff Schaller

35.1k952115

answered 3 hours ago

Jeff Schaller

35.1k952115

answered 3 hours ago

Jeff Schaller

35.1k952115

answered 3 hours ago

Jeff Schaller

35.1k952115

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

add a commentÂ |Â

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

what actually declare -A do?
â€“Â yael
2 hours ago

@yael, it declares an associative array
â€“Â Jeff Schaller
2 hours ago

add a commentÂ |Â

up vote
3
down vote

Look at what happens to the array when you initialize it:

$ i=0; for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -a A=([0]="3")

Only one element is present, and it has the index zero.

$ a=123; b=456; c=789; d=999; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done;
$ declare -p A
declare -a A=([123]="0" [456]="1" [789]="2" [999]="3")

To actually use the strings themselves as indices, declare the array as an associative array first with declare -A arrayname or typeset -A arrayname:

$ unset A; declare -A A; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -A A=([a]="0" [b]="1" [c]="2" [d]="3" )

answered 3 hours ago

ilkkachu

53.3k780146

add a commentÂ |Â

up vote
3
down vote

Look at what happens to the array when you initialize it:

$ i=0; for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -a A=([0]="3")

Only one element is present, and it has the index zero.

$ a=123; b=456; c=789; d=999; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done;
$ declare -p A
declare -a A=([123]="0" [456]="1" [789]="2" [999]="3")

To actually use the strings themselves as indices, declare the array as an associative array first with declare -A arrayname or typeset -A arrayname:

$ unset A; declare -A A; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -A A=([a]="0" [b]="1" [c]="2" [d]="3" )

answered 3 hours ago

ilkkachu

53.3k780146

add a commentÂ |Â

up vote
3
down vote

Look at what happens to the array when you initialize it:

$ i=0; for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -a A=([0]="3")

Only one element is present, and it has the index zero.

$ a=123; b=456; c=789; d=999; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done;
$ declare -p A
declare -a A=([123]="0" [456]="1" [789]="2" [999]="3")

To actually use the strings themselves as indices, declare the array as an associative array first with declare -A arrayname or typeset -A arrayname:

$ unset A; declare -A A; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -A A=([a]="0" [b]="1" [c]="2" [d]="3" )

answered 3 hours ago

ilkkachu

53.3k780146

Look at what happens to the array when you initialize it:

$ i=0; for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -a A=([0]="3")

Only one element is present, and it has the index zero.

$ a=123; b=456; c=789; d=999; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done;
$ declare -p A
declare -a A=([123]="0" [456]="1" [789]="2" [999]="3")

To actually use the strings themselves as indices, declare the array as an associative array first with declare -A arrayname or typeset -A arrayname:

$ unset A; declare -A A; i=0;
$ for k in a b c d; do A[$k]=$((i++)); done; declare -p A
declare -A A=([a]="0" [b]="1" [c]="2" [d]="3" )

answered 3 hours ago

ilkkachu

53.3k780146

answered 3 hours ago

ilkkachu

53.3k780146

answered 3 hours ago

ilkkachu

53.3k780146

answered 3 hours ago

ilkkachu

53.3k780146

add a commentÂ |Â

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