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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