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bashexitsubshell

Terminate a bash script when assigning a function call into a variable

I have the following code to demonstrate:

#!/bin/bash
function demo() {

   echo "This is A"
   exit 1
   echo "This is B"
}


input=$(demo)
echo "${input}"
echo "This is C"

You can see if I run this script it will print out the following output:

This is A
This is C

Since I have exit 1 function, the script did not terminate and printed the last statement This is C

But when I call the function like this:

#!/bin/bash
function demo() {

   echo "This is A"
   exit 1
   echo "This is B"
}


demo
echo "${input}"
echo "This is C"

Then the program terminated and did not print the last statement "This is C". The output is:

This is A

Is there an explanation for this and how do I force the script to terminate when having the exit 1 there plus assigning the function to a variable like example given. Meaning that, no other statements should be printed after exit 1

Solution

  • You can check the exit status when assigning the variable, and exit the main script if it failed.

    input=$(demo) || exit 1