如果 function 按条件执行，则 Bash 数组无法填充

Question

我有一个 function 填充全局 bash 数组

#!/bin/bash

# Array to store debugger IDs
ID=()

fill_id()
{
    for i in 1 2 3 4 5
    do
        ID+=( $i )
    done

    echo "fill_ids      - entries in ID: ${#ID[*]}"
}

如果直接调用 function，则填充数组并将条目计数打印为 5

fill_id
echo "main          - entries in ID: ${#ID[*]}"

Output

fill_ids      - entries in ID: 5
main          - entries in ID: 5

但是if在条件中调用 function 则条目计数打印为 0

if (fill_id)
then
    echo "main          - entries in ID: ${#ID[*]}"
fi

Output

fill_ids      - entries in ID: 5
main          - entries in ID: 0

fill_fd的返回值也不影响任何东西。

Answer 1

you are executing function in a subshell so it changes variable of child shell which is not visible in parent shell culprit is "()" you can debug it with printing shell id - see bellow:

#!/bin/bash

# Array to store debugger IDs
ID=()

echo $BASHPID
fill_id()
{
    for i in 1 2 3 4 5
    do
        ID+=( $i )
    done
echo $BASHPID
    echo "fill_ids      - entries in ID: ${#ID[*]}"
}


if  (fill_id)
then
    echo "main          - entries in ID: ${#ID[*]}"
fi

当您直接调用 function （不带子外壳）时，它会正确更改您的变量。

像这样：

if fill_id
then
    echo "main          - entries in ID: ${#ID[*]}"
fi

Answer 2

只是不要在子shell中执行 function 。

if fill_id
then
    echo "main          - entries in ID: ${#ID[*]}"
fi