How to pass positional parameters to a function

Question

I was trying things with bash scripts. I made this simple script

#!/bin/bash

function myfun()
{
    for item in `seq 1 5`
    do
        echo "$item $1 $2"
    done
}
myfun

but no luck. If I change it like this as below, everything seems to be fine,

#!/bin/bash

a=$1
b=$2

function myfun()
{
    for item in `seq 1 5`
    do
        echo "$item $a $b"
    done
}
myfun

It looks like arguments (positional parameters) do not work inside function in shell. Am I doing any mistake? I am still learning things. So can you explain why is it so?

Answer 1

It's the function not the loop:

function myfun()
{
    for item in `seq 1 5`
    do
        echo "$item $1 $2"
    done
}

# Pass all of the script's parameters to the function,
# as if writing  myfun "$1" "$2" "$3"..
myfun "$@"