How to safely echo all arguments of a script?

user_694509 0
Viewed 937

I am writing a bash script that must echo all of its arguments, which is surprisingly difficult to do.

The naive implementation looks like this:

#!/bin/bash
echo "$@"

However, that fails with input such as:

> ./script.sh -n -e -v -e -r
-v -e -r>

How can I make this more robust, such that the above results in:

> ./script.sh -n -e -v -e -r
-n -e -v -e -r
>
echo bash
4 Answers

echo command's behavior may be different between systems. The safest way is to use printf:

printf '%s\n' "$*"

According to posix:

It is not possible to use echo portably across all POSIX systems unless both -n (as the first argument) and escape sequences are omitted.

The printf utility can be used portably to emulate any of the traditional behaviors of the echo utility ...

user_9000780

You can use printf as well :

#!/bin/bash
printf "%s\n" "$*"
user_21256710

Using printf instead of echo:

#!/bin/bash

printf "%s " "$@"
printf "\n"
user_1272590

This one adds a space in the beginning but is rather simple : 

#!/bin/bash
echo "" "$@"
user_3267300
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