BufferUnderflowException while trying to read an Int from a binary file in Java

Question

I am trying to read 4 bytes which represent an int, located at byte position 64 in a binary file.

This is what I have tried:

package testbinaryfile2;

import java.io.IOException;
import java.nio.ByteBuffer;
import java.nio.channels.FileChannel;
import java.nio.file.Paths;
import java.nio.file.StandardOpenOption;

public class TestBinaryFile2 {

    public static void main(String[] args) throws IOException {
        
            FileChannel fc;    
            ByteBuffer indexField = ByteBuffer.allocate(4);
            
            fc = (FileChannel.open(Paths.get("myBinaryFile.bin"), StandardOpenOption.READ));                        
            fc.position(64);
            fc.read(indexField);
            System.out.println(indexField.getInt());
            
    }
            
}

This is the error I get:

run:
Exception in thread "main" java.nio.BufferUnderflowException
    at java.nio.Buffer.nextGetIndex(Buffer.java:509)
    at java.nio.HeapByteBuffer.getInt(HeapByteBuffer.java:373)
    at testbinaryfile2.TestBinaryFile2.main(TestBinaryFile2.java:30)
/home/user/.cache/netbeans/11.3/executor-snippets/run.xml:111: The following error occurred while executing this line:
/home/user/.cache/netbeans/11.3/executor-snippets/run.xml:94: Java returned: 1
BUILD FAILED (total time: 0 seconds)

M.E. M.E. · Accepted Answer · 2020-11-14T12:47:34

After read indexField is pointing to its end, so it is necessary to use .rewind() method before getting the int.

This code works:

package testbinaryfile2;

import java.io.IOException;
import java.nio.ByteBuffer;
import java.nio.channels.FileChannel;
import java.nio.file.Paths;
import java.nio.file.StandardOpenOption;

public class TestBinaryFile2 {

    public static void main(String[] args) throws IOException {
        
            FileChannel fc;    
            ByteBuffer indexField = ByteBuffer.allocate(4);
            
            fc = (FileChannel.open(Paths.get("myBinaryFile.bin"), StandardOpenOption.READ));                        
            fc.position(64);
            fc.read(indexField);
            indexField.rewind();    // <-- ADD THIS
            System.out.println(indexField.getInt());
            
    }
            
}

Additionally I got a suggestion to use mapped buffer:

  public static void main(String[] args) throws Exception {
    FileChannel fc = (FileChannel.open(Paths.get("myBinaryFile.bin"), StandardOpenOption.READ));
    MappedByteBuffer buffer = fc.map(FileChannel.MapMode.READ_ONLY, 0, fc.size());
    buffer.position(64);
    System.out.println(buffer.getInt());
  }

But I am wondering if fc.map reads the whole file at once and uses as much memory as the size of the file.

BufferUnderflowException while trying to read an Int from a binary file in Java

1 Answers