What does "$@" do in a bash script? [duplicate]

I came across this script:

#!/bin/sh
qemu-system-x86_64 -enable-kvm \
                   -m 2G \
                   -device virtio-vga,virgl=on \
                   -drive file=/home/empty/qemubl/lab.img,format=raw,if=virtio \
                   -cpu host \
                   -smp 4 \
                   -soundhw sb16,es1370 \
                   "$@"

What is the role of $@?

I searched man bash for $@ but got Pattern not found.

Solution 1:

Your search may have yielded no result because (as steeldriver commented) you didn't escape the $ , try searching for \$@ and you'll find the following under man bash – PARAMETERS – Special Parameters:

@  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 (i.e., they are removed).

Unquoted it gets expanded to $1 $2 $3 … ${N}, quoted to "$1" "$2" "$3" … "${N}", which is what you want in most cases. This and the difference to $* is very good explained in the bash-hackers wiki.

Solution 2:

It's an internal variable which is used to forward all arguments passed to the script, to the command which the script is calling, in this case qemu-system-x86_64.