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The set command in the following loop is confusing to me.
for i in "$@"
do
set -- "$@" "$i" # what does it mean?
done
I can understand $@ is all the positional parameters, and $i is one of the positional parameters. However, I can't figure out what
set -- "$@" "$i"
means.
348
bash [filename] runs the commands saved in a file. $@ refers to all of a shell script's command-line arguments. $1 , $2 , etc., refer to the first command-line argument, the second command-line argument, etc. Place variables in quotes if the values might have spaces in them.
• $* - It stores complete set of positional parameter in a single string. • $@ - Quoted string treated as separate arguments. • $? - exit status of command.
There is no difference if you do not put $* or $@ in quotes. But if you put them inside quotes (which you should, as a general good practice), then $@ will pass your parameters as separate parameters, whereas $* will just pass all params as a single parameter.
set allows you to change the values of shell options and set the positional parameters, or to display the names and values of shell variables.
It's appending the value of $i onto the end of the positional parameters. Not sure why one would want to do it, but it's basically a verbose way of doubling the parameters. It has the same affect as
$ set -- a b c
$ echo "$@"
a b c
$ set -- "$@" "$@"
echo "$@"
a b c a b c
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