What does $* mean in bash scripting?

From the man page:

* 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.

So it is equivalent to all the positional parameters, with slightly different semantics depending on whether or not it is in quotes.

It's all the arguments passed to the script, except split by word. You almost always want to use "$@" instead. And it's all in the bash(1) man page.

Its the list of arguments supplied on the command line to the script .$0 will be the script name.

It's a space separated string of all arguments. For example, if $1 is "hello" and $2 is "world", then $* is "hello world". (Unless $IFS is set; then it's an $IFS separated string.)

You can use symbolhound search engine to find codes that google will not look for.

For your query click here

If you see $ in prefix with anything , it means its a variable. The value of the variable is used.

Example:

count=100
echo $count
echo "Count Value = $count"

Output of the above script:

100
Count Value = 100

As an independent command it doesn't have any significance in bash scripting. But, as per usage in commands, it's used to indicate common operation on files / folders with some common traits.

and with grep used to represent zero or more common traits in a command.

Well,such case when you find number of average or total number.

$# number of argument to pass in the number $* number of all argument to pass