如何正确打印作为参数传递给 Bash 函数的字符串？

Question

I'm writing shell(bash) script like this:

output_function()
{
    for i in "$@"
    do
        echo $i
    done
}

process_funtion()
{
    string=process some thing
    output_function $string
}

for example, after process some thing, string is

i am line 1
i am line 2

I want to print these 2 line as it is, but actually I got

i
am
line
1
i
am
line
2

Also this NOT work:

#!/bin/bash

output()
{
    printf '%s\n' "$@"
}

output `ifconfig`

result is:

...
2000
inet6
fe80::6de5:743c:addd:7c5a%utun0
prefixlen
64
scopeid
0xa
nd6
options=201<PERFORMNUD,DAD>

And this NOT work also:

#!/bin/bash

output()
{
    printf '%s\n' "$*"
}

output `ifconfig`

result is all in one line.

how to fix this? thank you~

Answer 1

You need to enclose the function arguments in double quotes to prevent word splitting :

output "$string"

instead of

output $string

And you don't really need a loop to print the contents of $@ , you could simply write:

printf '%s\n' "$@"

---

See also:

• When to wrap quotes around a shell variable?
• Word Splitting - Bash Reference Manual

Answer 2

In your example:

output()
{
    printf '%s\n' "$*"
}

output `ifconfig`

the result of ifconfig also needs to be quoted, otherwise the result will be split into multiple arguments (using $IFS) before being passed to the function. So

output "`ifconfig`"

should do.

See Bash Reference Manual: 3.4.2 Special Parameters for using $* and $@ correctly and the difference between both.