Question

Java

Any help is appreciated

8.What is the output of the following program?

import java.io.*;

public class Q8 {

public static void main (String [] args)throws IOException {

RandomAccessFile file = new RandomAccessFile (new File ("Q8.txt"), "rw");

double x = 100.00

;

file.writeDouble(x);

file.writeDouble(x+10);

file.writeDouble(x+20);

file.seek(8);

System.out.println(file.readDouble());

System.out.println(file.getFilePointer());

}

}

--------------------------------------------

Output

110.00

16

My Question: why is this the output?   

Answer 1

******************************************************************************************
Please Upvote the answer as it matters to me a lot :)
*****************************************************************************************
As per HomeworkLib expert answering guidelines,Experts are supposed to answer only certain number of questions/sub-parts in a post.Please raise the remaining as a new question as per HomeworkLib guidelines.
******************************************************************************************

seek() sets the fp(file-pointer) offset from the beginning of the file, so seek(8) will set the fp to 8.

Remember in java the size of double is 8 bytes and generally, systems will be byte-addressable, so the file had content

100.00 110.00 120.00 where the first 8 bytes (0-7)will have value 100.00 so the byte address (8-15) will have the value 110.00

so when you try to execute the statement System.out.println(file.readDouble()); it will print 110.00

so after reading the value 110.00 the fp now points to the address 16 since the value 110.00 is double and is of 8 bytes size

address value

0-7 100.00

8-15 110.00

16-23 120.00