[Solved] Given this script snippet:function greet { echo "Hello, $1" } name="Alice" greet $nameWhat will be the output and why? | Bash Scripting

Bash Scripting - Functions

Given this script snippet:

function greet {
  echo "Hello, $1"
}
name="Alice"
greet $name

What will be the output and why?

AHello, $1 because variables inside functions are not expanded

BHello, name because $name is treated as literal

CHello, Alice because variable is expanded before function call

DError due to missing parentheses in function definition

Step-by-Step Solution

Solution:

Step 1: Understand variable expansion in Bash
Variables like $name are expanded before the function call, so greet $name becomes greet Alice.
Step 2: Function uses $1 to print first argument
The function prints "Hello, $1", so it outputs "Hello, Alice".
Final Answer:
Hello, Alice because variable is expanded before function call -> Option C
Quick Check:
Variables expand before function call, $1 holds argument [OK]

Quick Trick: Variables expand before function call; use $1 inside function [OK]

Common Mistakes:

MISTAKES

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