Reg. RandomAccessFile

[Angela Narain]

Below is a question on IO from JavaCaps mock exam#1 :

What is the output displayed by the following code? 
import java.io.*; 
public class TestIPApp { 
       public static void main(String args[]) throws IOException { 
             RandomAccessFile file = new RandomAccessFile("test.txt", "rw"); 
             file.writeBoolean(true); 
             file.writeInt(123456); 
             file.writeInt(7890); 
             file.writeLong(1000000); 
             file.writeInt(777); 
             file.writeFloat(.0001f); 
             file.seek(5); 
             System.out.println(file.readInt()); 
             file.close(); 
      } 
} 
Select correct answer: 
A) 123456 
B) 7890 
C) 1000000 
D) .0001 

The correct answer is B)
I am not clear as to how the data will be stored, and how
the seek will locate position 5. I know that seek
is used to the next read operation and starts from index 0.
Can some one explain to me .
Thanks a lot

[Richard Barnes]

Hi,
The seek() method works in bytes. The boolean value at the start will take up 1 byte (even though a boolean is actually only 1 bit long) at position 0, then the first int will take up the next 4 bytes (postions 1,2,3 and 4). Position 5 would be the first byte of the second int and hence the second int is displayed.
Regards,
Richard

[Angela Narain]

Thanks Richard.