Assigning arrays to variables in Bash script seems rather complicated:
a=("a" "b" "c")
b=$a
echo ${a[0]}
echo ${a[1]}
echo ${b[0]}
echo ${b[1]}
leads to
a
b
a
instead of
a
b
a
b
Why? How can I fix it?
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2 Answers
If you want to copy a variable that holds an array to another name, you do it like this:
a=('a' 'b' 'c')
b=( "${a[@]}" )
4 Comments
Tino
Beware of the sideffects: This renumbers the keys of the array/assoc-array, see:
a=(); a[1]=x; a[3]=y; a[5]=z; b=("${a[@]}"); declare -p a b gives declare -A a='([1]="x" [3]="y" [5]="z" )'; declare -a b='([0]="x" [1]="y" [2]="z")' and a=([a]=x [b]=y [c]=z); b=("${a[@]}"); declare -p a b gives declare -A a='([a]="x" [b]="y" [c]="z" )'; declare -a b='([0]="x" [1]="y" [2]="z")' (; added for better readability)
mjs
also if a is empty then b will have an item with empty content.
Binarus
8 years later, but could you please explain what you mean by reassign? According to my tests and the manual, the array contents actually are copied.
"${a[@]}" expands to a list of words (one word for each array member), enclosing that list in ( ) makes for a suitable RHS which can be used in an assignment, and b=... finally carries out the assignment. I can't see how a could lose its contents during this, which the term reassign implies. Did I miss something?
kojiro
You didn't miss anything. I can understand why you have a more nuanced definition of reassign in mind, but I think OP (at least at the time) didn't care about the old name
a and just wanted to know how to get all the values from a into b.Why?
If a is an array, $a expands to the first element in the array. That is why b in your example only has one value. In bash, variables that refer to arrays aren't assignable like pointers would be in C++ or Java. Instead variables expand (as in Parameter Expansion) into strings and those strings are copied and associated with the variable being assigned.
How can I fix it?
To copy a sparse array that contains values with spaces, the array must be copied one element at a time by the indices - which can be obtained with ${!a[@]}.
declare -a b=()
for i in ${!a[@]}; do
b[$i]="${a[$i]}"
done
From the bash man page:
It is possible to obtain the keys (indices) of an array as well as the values. ${!name[@]} and ${!name[*]} expand to the indices assigned in array variable name. The treatment when in double quotes is similar to the expansion of the special parameters @ and * within double quotes.
Here's a script you can test on your own:
#!/bin/bash
declare -a a=();
a[1]='red hat'
a[3]='fedora core'
declare -a b=();
# Copy method that works for sparse arrays with spaces in the values.
for i in ${!a[@]}; do
b[$i]="${a[$i]}"
done
# does not work, but as LeVar Burton says ...
#b=("${a[@]}")
echo a indicies: ${!a[@]}
echo b indicies: ${!b[@]}
echo "values in b:"
for u in "${b[@]}"; do
echo $u
done
Prints:
a indicies: 1 3
b indicies: 1 3 # or 0 1 with line uncommented
values in b:
red hat
fedora core
This also works for associative arrays in bash 4, if you use declare -A (with capital A instead of lower case) when declaring the arrays.
3 Comments
konsolebox
b=( "${a[@]}" ) copies elements with spaces. See Special Parameters
Chad Skeeters
Yes, but it doesn't support sparse arrays.
Ed Morton
This one got me too - I imagined a sparse array as meaning missing values,
unset a; a=( x '' '' y ), but on a re-read I now see "sparse" in this context is being used to mean missing indices unset a; a[0]=x; a[3]=y. Use b=( "${a[@]}" ); declare -p a b after each of those a initializations to see the difference. The solution copies array values, not whole arrays.