从文件中读取二进制数，并从longs / doubles中读取整数

Question

In this programmer i found Prime numbers in first 100.Numbers are in INT format and totally number of them is in DOUBLE format.I want to read that file and i did it for only INT numbers but i dont know hot to do it for DOUBLE number. Here is the code:

package int1;

import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.nio.ByteBuffer;
import java.nio.DoubleBuffer;
import java.nio.IntBuffer;
import java.nio.LongBuffer;
import java.nio.channels.FileChannel;

public class int_upis {
public static void main(String[] args) {
    File a = new File("C:\\Users\\Jovan\\Desktop\\Pisem.txt");
    FileOutputStream fos = null;
    try {
        fos = new FileOutputStream(a);
    } catch (Exception e) {

    }
    FileChannel ch = fos.getChannel();
    ByteBuffer bff = ByteBuffer.allocate(100);
    IntBuffer ibf = bff.asIntBuffer(); // Int type
    DoubleBuffer db = bff.asDoubleBuffer(); // Double type
    double p = 0;
    for (int i = 1; i <= 100; i++) {
        int t = 0;
        for (int j = 1; j <= i; j++) {
            if (i % j == 0) {
                t = t + 1;
            }

        }
        if (t < 3) {
            p = p + 1; // number of Prime numbers
            System.out.println(i);
            ibf.put(i);
            bff.position(4 * ibf.position());
            bff.flip();
            try {
                ch.write(bff);
                bff.clear();
                ibf.clear();
            } catch (IOException e) {

            }
        }

    }

    try {
        db.put(p); //At the end of the txt-file i put double format of number (Number of Prime numbers)
        bff.position(8*db.position());
        bff.flip();
        ch.write(bff);

        System.out.println("File is writen with: " + ch.size());
        ch.close();
    } catch (IOException e) {
        e.printStackTrace();
    }

}

}

Now I tried to read this file:

public class int_ispis {
public static void main(String[] args) throws IOException {
    File a = new File("C:\\Users\\Jovan\\Desktop\\Pisem.txt");
    FileInputStream fis = null;
    try {
        fis = new FileInputStream(a);
    } catch (Exception e) {

    }
    FileChannel ch = fis.getChannel();
    ByteBuffer bff = ByteBuffer.allocate(6 * 4);

This is one line of Prime Number put in a 6-row array (this line below):

    int[] niz = new int[6];
    System.out.println("Pre flipa: " + bff.position() + " " + bff.limit());
    System.out.println("++++++++++++++++++++++++++++++++++++");
    while (ch.read(bff) != -1) {

        bff.flip();
        System.out.println();
        System.out.println("Posle flipa: " + bff.position() + " " + bff.limit());
        IntBuffer ib = bff.asIntBuffer();
        System.out.println("IB: " + ib.position() + " " + ib.limit());

        int read = ib.remaining();
        System.out.println(read);

When it come to the end of file it puts Double Number as Integer and writes wrong number(How to separate Integer form Double number?)

        ib.get(niz, 0, ib.remaining());
        for (int i = 0; i < read; i++) {
            System.out.print(niz[i] + " ");

        }
        System.out.println();
        System.out.println("=================================");
        ib.clear();
        bff.clear();

    }
}
}

Answer 1

A binary file does not have any "separators".

You need to know the structure of the file content and use this knowledge.

In this programmer i found Prime numbers in first 100.Numbers are in INT format and totally number of them is in DOUBLE format.

This means that there is only one long value in the file and this is in the last 8 bytes . So you simply have to check if the current position is fileLenght - 8 and then read these last 8 bytes as a long value.