How do i define arguments in bash functions?

giraffelife

Let's see the following script run, the idea is to see what variable get defined within the function and what get's defined outside.

$ cat bash.variables 

cargo()
{
  echo "++++++++++++++++++++++"
  echo "I'm inside a function, let me out: $0 $1 $2"
  var1="Flashy speakers"
  echo variable inside the function is: $var1
  echo "++++++++++++++++++++++"
}

var1="I've got my cargo on my ship"
echo "what's on my ship: $var1"
echo $0: $1 $2

cargo functionarg1 functionarg2
echo "what's on my ship: $var1"
echo $0: $1 $2

$ ./bash.variables cameras radios
what's on my ship: I've got my cargo on my ship
./bash.variables: cameras radios
++++++++++++++++++++++
I'm inside a function, let me out: ./bash.variables functionarg1 functionarg2
variable inside the function is: Flashy speakers
++++++++++++++++++++++
what's on my ship: Flashy speakers
./bash.variables: cameras radios

The variable inside the function are global, and the var1 that was defined outside the function is replaced with the value of the "var1" in the function.

Note: There is a statement "local" that can be used to keep the variable local to a function.