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bash array assignment
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Nicholas Geovanis
Feb 9, 2016, 8:24:41 PM2/9/16
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How does one assign one array to another in bash? Normal assignment seems not to produce an array, just a string. See this code fragment for example:
[ngeo@localhost src]$ cat arraytst1.sh
#!/usr/bin/bash
names2=()
names=("Bob" "Peter" "$USER" "Big Bad John")
names2="${names[@]}"
for name in "${names[@]}"; do echo "$name"; done
for name in "${names2[@]}"; do echo "$name"; done
[ngeo@localhost src]$ ./arraytst1.sh
Bob
Peter
ngeo
Big Bad John
Bob Peter ngeo Big Bad John
[ngeo@localhost src]$ cat arraytst1.sh
#!/usr/bin/bash
names2=()
names=("Bob" "Peter" "$USER" "Big Bad John")
names2="${names[@]}"
for name in "${names[@]}"; do echo "$name"; done
for name in "${names2[@]}"; do echo "$name"; done
[ngeo@localhost src]$ ./arraytst1.sh
Bob
Peter
ngeo
Big Bad John
Bob Peter ngeo Big Bad John
Janis Papanagnou
Feb 9, 2016, 8:37:24 PM2/9/16
to
On 10.02.2016 02:24, Nicholas Geovanis wrote:
> How does one assign one array to another in bash? Normal assignment seems
> not to produce an array, just a string. See this code fragment for
> example:
>
> [ngeo@localhost src]$ cat arraytst1.sh
> #!/usr/bin/bash
> names2=()
> names=("Bob" "Peter" "$USER" "Big Bad John")
> names2="${names[@]}"
names2=( "${names[@]}" )
> How does one assign one array to another in bash? Normal assignment seems
> not to produce an array, just a string. See this code fragment for
> example:
>
> [ngeo@localhost src]$ cat arraytst1.sh
> #!/usr/bin/bash
> names2=()
> names=("Bob" "Peter" "$USER" "Big Bad John")
> names2="${names[@]}"
Janis
Nicholas Geovanis
Feb 9, 2016, 8:44:30 PM2/9/16
to
On Tuesday, February 9, 2016 at 7:24:41 PM UTC-6, Nicholas Geovanis wrote:
> How does one assign one array to another in bash? Normal assignment seems not > to produce an array, just a string. See this code fragment for example:
Managed to answer my own question, See below. Apologies for wasted bandwidth.
> How does one assign one array to another in bash? Normal assignment seems not > to produce an array, just a string. See this code fragment for example:
[ngeo@localhost src]$ cat arraytst1.sh
#!/usr/bin/bash
#names2=()
#!/usr/bin/bash
names=("Bob" "Peter" "$USER" "Big Bad John")
names2=("${names[@]}")
Stephane Chazelas
Feb 10, 2016, 5:40:14 PM2/10/16
to
2016-02-10 02:37:20 +0100, Janis Papanagnou:
Note that it only works for non-sparse arrays. In the general
case, that would mess up the indices.
$ bash -c 'a=([12]=a [23]=b [34]=c); b=("${a[@]}"); declare -p a b'
declare -a a='([12]="a" [23]="b" [34]="c")'
declare -a b='([0]="a" [1]="b" [2]="c")'
To copy an array in bash, I think you need a loop:
b=()
for i in "${!a[@]}"; do
b[i]=${a[i]}
done
That's the same situation as ksh.
Arrays of all other shells (rc, es, csh, tcsh, zsh) are normal
arrays, not sparse, so the situation is a lot simpler there.
--
Stephane
> On 10.02.2016 02:24, Nicholas Geovanis wrote:
> > How does one assign one array to another in bash? Normal assignment seems
> > not to produce an array, just a string. See this code fragment for
> > example:
> >
> > [ngeo@localhost src]$ cat arraytst1.sh
> > #!/usr/bin/bash
> > names2=()
> > names=("Bob" "Peter" "$USER" "Big Bad John")
> > names2="${names[@]}"
>
> names2=( "${names[@]}" )
[...]
> > How does one assign one array to another in bash? Normal assignment seems
> > not to produce an array, just a string. See this code fragment for
> > example:
> >
> > [ngeo@localhost src]$ cat arraytst1.sh
> > #!/usr/bin/bash
> > names2=()
> > names=("Bob" "Peter" "$USER" "Big Bad John")
> > names2="${names[@]}"
>
> names2=( "${names[@]}" )
Note that it only works for non-sparse arrays. In the general
case, that would mess up the indices.
$ bash -c 'a=([12]=a [23]=b [34]=c); b=("${a[@]}"); declare -p a b'
declare -a a='([12]="a" [23]="b" [34]="c")'
declare -a b='([0]="a" [1]="b" [2]="c")'
To copy an array in bash, I think you need a loop:
b=()
for i in "${!a[@]}"; do
b[i]=${a[i]}
done
That's the same situation as ksh.
Arrays of all other shells (rc, es, csh, tcsh, zsh) are normal
arrays, not sparse, so the situation is a lot simpler there.
--
Stephane
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