You want to write a Bash function that increments a counter variable without affecting the global variable of the same name. Which approach correctly uses local to achieve this?

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Bash Scripting - Functions

You want to write a Bash function that increments a counter variable without affecting the global variable of the same name. Which approach correctly uses local to achieve this?

AIncrement the global variable directly without declaring local.

BDeclare <code>local counter</code> inside the function and increment it there.

CUse <code>export counter</code> inside the function before incrementing.

DDeclare <code>counter</code> outside the function and use <code>local</code> globally.

Step-by-Step Solution

Solution:

Step 1: Understand local variable purpose
Declaring local counter inside the function creates a separate variable that does not affect the global one.
Step 2: Check other options
Incrementing global directly changes global variable; export shares variables with child processes; declaring local outside function is invalid.
Final Answer:
Declare local counter inside the function and increment it there. -> Option B
Quick Check:
Local inside function isolates variable changes [OK]

Quick Trick: Use 'local' inside function to isolate variable changes [OK]

Common Mistakes:

MISTAKES

Modifying global variable unintentionally
Misusing export for local scope
Declaring local variables outside functions

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More Bash Scripting Quizzes

You want to write a Bash function that increments a counter variable without affecting the global variable of the same name. Which approach correctly uses local to achieve this?

Step 1: Understand local variable purpose

Step 2: Check other options

Final Answer:

Quick Check:

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