shell命令末尾的“$ *”是什么意思

Question

In a sample shell script, the command was -

. <sourced_file.sh> $*

What does the $* mean? Thanks.

Answer 1

$* expands to all of the arguments that were given to the script in which it appears, or to the current shell function if it appears inside a function.

It's usually incorrect usage though, because it breaks arguments that contain spaces into multiple arguments. More correct is "$@" which preserves the original arguments even if they have spaces in them.

Answer 2

$* is a variable holding all positional parameters starting from 1 (the arguments to the current shell script)

man 1 bash :

The shell treats several parameters specially. These parameters may only be referenced; assignment to them is not allowed.

*

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.

@

Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (ie, they are removed).

Usually you want to use "$@" though:

"$*" is equivalent to "$1 $2 ..." whereas
"$@" is equivalent to "$1" "$2" ...

Answer 3

$* is the alias for all the arguments given to the current script.

For example, if you launch :

./test_script.sh arg1 arg2 arg3

If you do echo $* in test_script.sh you will display : arg1 arg2 arg3

Answer 4

Google on bash $* gives you immediately the advanced bash scripting guide which gives you the answer. You often should prefer "$@"

Answer 5

$* are the arguments

So if you script "foo.sh" is

#!/bin/sh
gethostip $*

And you call foo.sh -d localhost

it will be the same as

gethostip -d localhost

Minbd though, $* will not play nice with arguments having spaces.

Answer 6

It's “all the arguments”:

$ cat >test<<EOF
> echo \$*
> EOF
$ bash test foo bar
foo bar