Re: Prompting \n

T o n g wrote:
What I didn't make clear in the OP was that, I was actually looking for a
general purpose command that can show whatever messages thrown to it,
including the common backslash escapes. echo -e does a good job, but

But 'echo -e' isn't portable.

printf will choke on any of its control characters, e.g., %.

In that case use %b and print everything as a string like echo.

$ printf "%b" "Hello world!"

$ printf "%b" "Hello %1 %2 %3"
Hello %1 %2 %3

$ printf "%b" "Hello1 %1\nHello2 %2\nHello3 %3"
Hello1 %1
Hello2 %2
Hello3 %3

Today, I just found that printf cannot be used to show whatever the
command line is.

Yes it can.

Here is an example:

set -- command -opt param params...

echo='echo -e'

$ $echo "\n> $@\n"

> command -opt param params...

That is quite a convoluted example.

So far so good. Now try printf:

$ printf "\n> $@\n"

> command

What's happening? why printf chokes on '-'?

Because $@ is quoted it is being split up into separate arguments. If
you want to use all of the arguments together in a string you need to
use $* not $@. This works as you expect.

$ printf "\n> $*\n"

You can see what "$@" does by using it with a for loop and printing
each argument in turn.

$ for i in "\n> $@\n"; do echo _"$i"_;done
_\n> command_
_-opt_
_param_
_params...\n_

$ for i in "\n> $*\n"; do echo _"$i"_;done
_\n> command -opt param params...\n_

Use "$@" when you want each argument in separate strings. Use "$*"
when you want all of the arguments together in one string.

BTW, the reason that I gave up 'echo -e' was that it started to
mysteriously output that "-e " in front of the messages I wanted to show
in my /bin/sh scripts. I still haven't figure out why yet.

You were probably calling it twice with echo -e -e.

Bob

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