[Solved] Find the problem in this Java program snippet:public class Sample { public static void main(String[] args) { for(int i=1; i | Java

Java - Command Line Arguments

Find the problem in this Java program snippet:

public class Sample {
  public static void main(String[] args) {
    for(int i=1; i<=args.length; i++) {
      System.out.println(args[i]);
    }
  }
}

ANo problem, code runs fine

BLoop index should start at 0, not 1

CSystem.out.println cannot print args[i]

Dargs.length should be args.size()

Step-by-Step Solution

Solution:

Step 1: Understand array indexing in Java
Arrays start at index 0, so loop should start at 0 to access first element.
Step 2: Check loop bounds
Loop runs i from 1 to args.length inclusive, causing ArrayIndexOutOfBoundsException at last iteration.
Final Answer:
Loop index should start at 0, not 1 -> Option B
Quick Check:
Array indexing starts at 0 in Java [OK]

Quick Trick: Array indexes start at 0, loops must reflect that [OK]

Common Mistakes:

Starting loop at 1
Using args.size() instead of length
Assuming print can't handle args[i]

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More Java Quizzes

Find the problem in this Java program snippet:

Step 1: Understand array indexing in Java

Step 2: Check loop bounds

Final Answer:

Quick Check:

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