[Solved] Consider this script:#!/bin/bash count=0 increment() { local count=$((count + 1)) echo $count } increment increment echo $countWhat will be the output and why? | Bash Scripting

Bash Scripting - Functions

Consider this script:

#!/bin/bash
count=0
increment() {
  local count=$((count + 1))
  echo $count
}
increment
increment
echo $count

What will be the output and why?

A1 2 2 because local count updates global count

B1 1 0 because local count shadows global count inside function

C0 0 0 because local count is not incremented

D1 2 0 because local count resets each call

Step-by-Step Solution

Solution:

Step 1: Analyze local variable initialization each call
Each call initializes local count as $((count + 1)), but count inside is local and uninitialized, so it uses global count which is 0, so local count becomes 1 each time.
Step 2: Check output of each call and global count
Each call prints 1, but since local count resets, second call also prints 1. Global count remains 0.
Step 3: Verify options
Only 1 1 0 because local count shadows global count inside function matches output 1 1 0 and explains that local count shadows global count inside function.
Final Answer:
1 1 0 because local count shadows global count inside function -> Option B
Quick Check:
Local variables reset each call = A [OK]

Quick Trick: Local variables reset each call, don't accumulate [OK]

Common Mistakes:

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