将元素附加到 bash 中的数组

Question

I tried the += operator to append an array in bash but do not know why it did not work

#!/bin/bash


i=0
args=()
while [ $i -lt 5 ]; do

    args+=("${i}")
    echo "${args}"
    let i=i+1

done

expected results

0
0 1
0 1 2
0 1 2 3
0 1 2 3 4

actual results

0
0
0
0
0

Any help would be appreciated.

Answer 1

It did work, but you're only echoing the first element of the array. Use this instead:

echo "${args[@]}"

Bash's syntax for arrays is confusing. Use ${args[@]} to get all the elements of the array. Using ${args} is equivalent to ${args[0]} , which gets the first element (at index 0).

See ShellCheck : Expanding an array without an index only gives the first element.

Also btw you can simplify let i=i+1 to ((i++)) , but it's even simpler to use a C-style for loop. And also you don't need to define args before adding to it.

So:

#!/bin/bash
for ((i=0; i<5; ++i)); do
    args+=($i)
    echo "${args[@]}"
done