How to pass arguments from one function to another function in bash scripting?

I can do this in Python:

def one(arg1): return arg1
def two(a,b): result=a+b return one(result)
two(1,3)

And it will work. But how do I do the same in a bash script?

1 Answer

Try that argument passing this way:

#!/usr/bin/env bash
function one(){ # Print the result to stdout echo "$1"
}
function two() { local one=$1 local two=$2 # Do arithmetic and assign the result to # a variable named result result=$((one + two)) # Pass the result of the arithmetic to # the function "one" above and catch it # in the variable $1 one "$result"
}
# Call the function "two"
two 1 3