Java基础之写文件——缓冲区中的多条记录(PrimesToFile3)

控制台程序,上一条博文(PrimesToFile2)每次将一个素数写入到文件中,所以效率不是很高。最好是使用更大的缓冲区并加载多个素数。

本例重复使用三个不同的视图缓冲区加载字节缓冲区并尽可能加入更多的素数,推荐使用这种方式。

 1 import static java.lang.Math.ceil;
 2 import static java.lang.Math.sqrt;
 3 import static java.lang.Math.min;
 4 import static java.nio.file.StandardOpenOption.*;
 5 import java.nio.file.*;
 6 import java.nio.channels.*;
 7 import java.nio.*;
 8 import java.util.*;
 9 import java.io.IOException;
10 
11 public class PrimesToFile3 {
12   public static void main(String[] args) {
13     int primesRequired = 100;                                          // Default count
14     if (args.length > 0) {
15       try {
16         primesRequired = Integer.valueOf(args[0]).intValue();
17       } catch (NumberFormatException e) {
18         System.out.println("Prime count value invalid. Using default of " + primesRequired);
19       }
20     }
21 
22       long[] primes = new long[primesRequired];                        // Array to store primes
23       
24       getPrimes(primes);
25       Path file = createFilePath("Beginning Java Struff","primesAgain.txt");
26       writePrimesFile(primes,file);
27     }
28       //Calculate enough primes to fill the array
29       private static long[] getPrimes(long[] primes) {
30           primes[0] = 2L;                                                  // Seed the first prime
31           primes[1] = 3L;                                                  // and the second
32           // Count of primes found ?up to now, which is also the array index
33           int count = 2;
34           // Next integer to be tested
35           long number = 5L;
36 
37           outer:
38           for (; count < primes.length; number += 2) {
39 
40             // The maximum divisor we need to try is square root of number
41             long limit = (long)ceil(sqrt((double)number));
42 
43             // Divide by all the primes we have up to limit
44             for (int i = 1 ; i < count && primes[i] <= limit ; ++i)
45               if (number % primes[i] == 0L)                                // Is it an exact divisor?
46                 continue outer;                                            // yes, try the next number
47 
48             primes[count++] = number;                                      // We got one!
49           }
50           return primes;
51         }
52     //Create the path for the named file in the specified directory
53     //in the user home directory
54     private static Path createFilePath(String directory, String fileName) {
55         Path file = Paths.get(System.getProperty("user.home")).resolve(directory).resolve(fileName);
56         try {
57           Files.createDirectories(file.getParent());                       // Make sure we have the directory
58         } catch (IOException e) {
59           e.printStackTrace();
60           System.exit(1);
61         }
62         System.out.println("New file is: " + file);
63         return file;
64     }
65     
66     //Write the array contents to file
67     private static void writePrimesFile(long[] primes, Path file) {
68         final int BUFFERSIZE = 1024;                                        // Byte buffer size
69         try (WritableByteChannel channel = Files.newByteChannel( file, EnumSet.of(WRITE, CREATE))){
70           ByteBuffer buf = ByteBuffer.allocate(BUFFERSIZE);
71           String primeStr = null;
72           int primesWritten = 0;
73         while (primesWritten < primes.length) {
74             while (primesWritten < primes.length) {
75               primeStr = "prime = " + primes[primesWritten];
76               if ((buf.position() + 2*primeStr.length() + 16) > buf.limit()) {
77                 break;
78               }
79               buf.asDoubleBuffer().put(0, (double)primeStr.length());
80               buf.position(buf.position() + 8);
81               buf.position(buf.position() + 2*buf.asCharBuffer().put(primeStr).position());
82               buf.asLongBuffer().put(primes[primesWritten++]);
83               buf.position(buf.position() + 8);
84             }
85             buf.flip();
86             channel.write(buf);
87             buf.clear();
88           }
89          System.out.println("File written is " + ((FileChannel)channel).size() + " bytes.");
90         } catch (IOException e) {
91           e.printStackTrace();
92         }
93     }
94 }
原文地址:https://www.cnblogs.com/mannixiang/p/3386796.html