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Accepted
Atia
Asked: 2018-08-14 02:25:01 +0800 CST

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

  • 772

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?

bash functions

1 Answers

  1. Best 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
    • 2