How to randomly select a function and then execute it?

Question

bash

There are three functions:

function1() {
    echo пукпук
}
function2() {
    echo пукпук2
}
function3() {
    echo пукпук3
}

How can you randomly select a feature?

In python, this is done like this:

from random import choice

def func1():
  print("пукпук")

def func2():
  print("пукпук2")	

def func3():
  print("пукпук3")

getrandom = choice([func1, func2, func3])
getrandom()

How to do it in bash?

Answer 1

getRandom()
{
   case $(($RANDOM % 3)) in
   0) function1 ;;
   1) function2 ;;
   2) function3 ;;
   esac
}
echo `getRandom`

Answer 2

"function$((1 + ($RANDOM % 3)))"
The built-in function in bash $RANDOMwill return a random 16-bit integer from 0 to 32767 on each call. You can take the remainder of dividing it by 3 - it will be 0, 1 or 2 - and adding one to stick to the word "function" - you get the name of the desired function. It remains to call it - just this line in the bash script, in fact, calls the resulting function.

Checks

$ ./rand.sh 
пукпук
$ ./rand.sh 
пукпук
$ ./rand.sh 
пукпук3
$ ./rand.sh 
пукпук3
$ ./rand.sh 
пукпук3
$ ./rand.sh 
пукпук3
$ ./rand.sh 
пукпук3
$ ./rand.sh 
пукпук2
$ ./rand.sh 
пукпук2
$ ./rand.sh 
пукпук2
$ ./rand.sh 
пукпук2
$ ./rand.sh 
пукпук
$ cat ./rand.sh 
#!/bin/bash

function1() {
    echo пукпук
}
function2() {
    echo пукпук2
}
function3() {
    echo пукпук3
}

"function$((1 + ($RANDOM % 3)))"

Answer 3

eval
eval "function$((RANDOM%3+1))"
or directly
"function$((RANDOM%3+1))"