Issue with copying bash array and only first element being present

Question

I would like to declare an array, temporally change it's value while storing the old value, perform some computation and then later restore the original array value. Of this process, my original array declaration, reassigning the array, and even the temporary array copy seem to all be working. However I am running into an issue with the final restoration of the array back to its original value.

The below script demonstrates the issue. MY_ARR is the main array and SAVE_MY_ARR is the copy I am making while I modify the array temporarily before restoring it. As per the script comments, everything prints out as expected, with the exception of the original array after restoring its value. What am I doing incorrectly here?

The script:

IFS=$'\n'

MY_ARR=(one two three)
# GOOD: prints "one,two,three,"
for i in ${MY_ARR[@]}; do
    echo -n "$i,"
done
echo ""

SAVE_MY_ARR=("${MY_ARR[@]}")
MY_ARR=(foo)
# GOOD: prints "foo,"
for i in ${MY_ARR[@]}; do
    echo -n "$i,"
done
echo ""

# GOOD: prints "one;two;three"
for i in ${SAVE_MY_ARR[@]}; do
    echo -n "$i;"
done
echo ""

# BAD: prints "one," and not "one,two,three,"
MY_ARR=("${SAVE_MY_ARR}")
for i in ${MY_ARR[@]}; do
    echo -n "$i,"
done
echo ""

The output:

one,two,three,
foo,
one;two;three;
one,

Why don't you do the same thing that you did when you first saved the array? — Mat
– Mat, Commented May 20, 2014 at 12:59

GHugo · Accepted Answer · 2014-05-20 13:00:29Z

1

I think it's a typo mistake, here replace:

# BAD: prints "one," and not "one,two,three,"
MY_ARR=("${SAVE_MY_ARR}")

By

# BAD: prints "one," and not "one,two,three,"
MY_ARR=("${SAVE_MY_ARR[@]}")

Adding the [@] to SAVE_MY_ARR.

answered May 20, 2014 at 13:00

GHugo

2,70417 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Bryce Thomas Over a year ago

A careless mistake. Thanks for picking this up - it seems to fix the problem.

chepner · Accepted Answer · 2014-05-20 13:08:09Z

0

bash arrays are not first-class objects. In your case, you discovered that $foo does not expand to all the elements of the array foo; it is identical to ${foo[0]} and expands to only the first element of the array. The only way to copy an array is to create a new array which is initialized to the values of the first array, by expanding the first array into a sequence of its elements.

answered May 20, 2014 at 13:08

chepner

538k77 gold badges594 silver badges746 bronze badges

Collectives™ on Stack Overflow

Issue with copying bash array and only first element being present

2 Answers 2

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related