How to pass parameters to function in a bash script?

Question

I'd like to write a function that I can call from a script with many different variables. For some reasons I'm having a lot of trouble doing this. Examples I've read always just use a global variable but that wouldn't make my code much more readable as far as I can see.

Intended usage example:

#!/bin/bash #myscript.sh var1=$1 var2=$2 var3=$3 var4=$4 add(){ result=$para1 + $para2 } add $var1 $var2 add $var3 $var4 # end of the script ./myscript.sh 1 2 3 4 

I tried using $1 and such in the function, but then it just takes the global one the whole script was called from. Basically what I'm looking for is something like $1, $2 and so on but in the local context of a function. Like you know, functions work in any proper language.

Answer 1

To call a function with arguments:

function_name "$arg1" "$arg2" 

The function refers to passed arguments by their position (not by name), that is $1, $2, and so forth. $0 is the name of the script itself.

Example:

#!/bin/bash add() { result=$(($1 + $2)) echo "Result is: $result" } add 1 2 

Output

./script.sh Result is: 3 

Answer 2

In the main script $1, $2, is representing the variables as you already know. In the subscripts or functions, the $1 and $2 will represent the parameters, passed to the functions, as internal (local) variables for this subscripts.

#!/bin/bash #myscript.sh var1=$1 var2=$2 var3=$3 var4=$4 add(){ #Note the $1 and $2 variables here are not the same of the #main script... echo "The first argument to this function is $1" echo "The second argument to this function is $2" result=$(($1+$2)) echo $result } add $var1 $var2 add $var3 $var4 # end of the script ./myscript.sh 1 2 3 4