Using echo to print arguments inside a function

Shell Programming and Scripting
1

I am using echo in bash. Have created a function prargv which takes a number of arguments.

Example:

prargv "-e" "--examples"

Inside prargv, I want to print all the arguments using echo

echo "$@"

This returns

--examples

rather than

-e --examples"

This problem can be fixed by introducing a blank space

echo " $@"

How can I avoid echo interpreting -e when I do not put a blank at the beginning?

2

Hi.

As seen in man echo, for some systems -e is treated as an option:

       -e     enable interpretation of backslash escapes

Using printf as in:

printf "%s\n" "-e --example"

produces:

-e --example

This is for:

OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian 5.0.8 (lenny, workstation) 
printf - is a shell builtin [bash]

This also works with:

/usr/bin/printf "%s\n" "-e --example"

Best wishes ... cheers, drl

4

That is correct, the -e is treated as an option. I found it strange that echo is interpreting it as I put $@ in double quotes.

5

Hi.

An illustration of $* and $@:

#!/usr/bin/env bash

echo
echo " asterisk"
echo "$*"

echo
echo " at sign"
echo "$@"

echo
echo " asterisk"
printf "%s\n" "$*"

echo
echo " at sign"
printf "%s\n" "$@"

producing:

$ ./s1 -e --example

 asterisk
-e --example

 at sign
--example

 asterisk
-e --example

 at sign
-e
--example

See man pages for details, and experiment.

Best wishes ... cheers, drl