JAVA PROGRAM • LEVEL 14 — FILE HANDLING & SERIALIZATION

Read a File with BufferedReader in Java

Learn read a file with bufferedreader in java with a complete, compiler-ready Java 17 example.

IntermediateBufferedReadertry-with-resources

PROBLEM UNDERSTANDING

Input and expected output

Sample input
No input required
Sample output
Basics
Loops
Methods

COMPLETE JAVA 17 PROGRAM

Complete Java 17 implementation

java-buffered-file-reading.java
Open in compiler
import java.io.*;import java.nio.file.*;
class Main{public static void main(String[] args)throws Exception{Path path=Path.of("topics.txt");Files.writeString(path,"Basics\nLoops\nMethods\n");try(BufferedReader reader=Files.newBufferedReader(path)){String line;while((line=reader.readLine())!=null)System.out.println(line);}}}

GUIDED CODE TOUR • NOT LIVE EXECUTION

Study the program line by line

Use the real compiler button above to run and debug with different inputs.

CURRENT STEP

Select Start to walk through the important lines.

SELECTED LINE

No line selected

EXPECTED OUTPUT FOR THE SAMPLE

Basics
Loops
Methods
0%Step 0 of 0

PROGRAM EXPLANATION

Algorithm and explanation

  1. Read the sample input and convert it into the Java values required for read a file with bufferedreader in java.
  2. Apply BufferedReader and try-with-resources in the order shown by the program.
  3. Print the final result and compare it with the documented sample output.

This Java 17 example demonstrates BufferedReader and try-with-resources through a complete, executable program. The values are intentionally small so students can trace each statement, verify the output, and then modify the example safely in the online compiler.

EFFICIENCY

Time and space complexity

Time complexity

O(n)

Auxiliary space

O(1)

DEBUGGING CHECKLIST

Common mistakes

Check this

Keep the class name Main and the entry method signature public static void main(String[] args) when using the online compiler.

Check this

Check the loop boundary and update expression carefully to avoid skipping the final value or creating an infinite loop.

Check this

Java is case-sensitive, and every statement that requires a semicolon must end with one.

Try it yourself

Practice: Run the program with the sample input, predict its output, and then test one boundary case of your own.