Java的randomaccessfile

Question

I am trying to read ints from a text file. This file has the following text:

3 1.9 a2 8 9
1 3 5.6 xx 7
7.2 abs 7  :+  -4
5
ds ds ds

I'm using the randomAccessFile with java. When I try to read the first int I get 857747758 (I wrote int number1 = inputStream.nextInt();, as well and others like it for the other ints on the txt file.) it told me that the length is 59. I am just wondering why its giving me this large number, and not just 3? Also, where would i move the file pointer after i read int 3? i know that it's starts at location 0, but i just need help figuring out how file point moves. Does it count empty spaces? I know that it converts them into binary.

import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.RandomAccessFile;


public class Program5 {
public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    RandomAccessFile inputStream = null;
    int n1, n2, n3, n4, n5, n6, n7, n8, n9;
    int sum = 0;

    System.out.println("Please enter the file name you wish to read from.");
    String file_name = keyboard.next();

    try {
        inputStream = new RandomAccessFile(file_name, "r");

        n1 = inputStream.readInt();
        // inputStream.seek(3);
        // n2 = inputStream.readInt();
        // n3 = inputStream.readInt();
        // n4 = inputStream.readInt();
        // n5 = inputStream.readInt();
        // n6 = inputStream.readInt();
        // n7 = inputStream.readInt();
        // n8 = inputStream.readInt();
        // n9 = inputStream.readInt();
        System.out.println(inputStream.length());
        System.out.println(inputStream.getFilePointer());
        System.out.println(n1);
        // System.out.println(n2);
        sum = n1; // n2; /* n3 + n4 + n5 + n6 + n7 + n8 + n9; */
        inputStream.close();
    } catch (FileNotFoundException e) {
        System.out.println("Problem opening up file" + file_name);
        System.exit(0);
    } catch (IOException ex) {
        ex.printStackTrace();
    }
    System.out.println("The sum of the numbers is: " + sum);

    System.out.println("End of program.");

}
 }

getting this

 Please enter the file name you wish to read from.
 inputP5.txt
 length at 59
 file pointer at 4
 n1 =857747758
 The sum of the numbers is: 857747758
 End of program.

Answer 1

I am trying to read ints from a text file.

Then you shouldn't be using a binary API.

I'm using the RandomAccessFile with Java. When I try to read the first int I get 857747758 (I wrote int number1 = inputStream.nextInt()

No you didn't. You used inputStream.readInt() . That's a binary API. readInt() reads a 4-byte binary integer in network byte order, just like it says in the Javadoc.

You should be using Scanner.nextInt() .

Answer 2

The readInt reads the 4bytes. So you won't get the output u wanted.

The java api https://docs.oracle.com/javase/7/docs/api/java/io/RandomAccessFile.html#readInt()

Reads a signed 32-bit integer from this file. This method reads 4 bytes from the file, starting at the current file pointer. If the bytes read, in order, are b1, b2, b3, and b4, where 0 <= b1, b2, b3, b4 <= 255, then the result is equal to: (b1 << 24) | (b2 << 16) + (b3 << 8) + b4

try (RandomAccessFile raf = new RandomAccessFile(new File(
        "filename.txt"), "r")) {

    byte[] bt = new byte[4];
    raf.read(bt);
    System.out.println(Arrays.toString(bt));
    System.out.println((bt[0] << 24) | (bt[1] << 16) + (bt[2] << 8) + bt[3]);

    raf.seek(0);
    System.out.println(raf.readInt());
} catch (IOException e) {
    e.printStackTrace();
}

Here is the output

[51, 32, 49, 46]
857747758
857747758