Java基础之写文件——使用多个视图缓冲区（PrimesToFile2）

控制台程序。本例将对应于每个素数的数据以三个连续数据项的形式写入：

1、以二进制值表示的字符串长度值（最好是整型，但本例使用double类型）；

2、素数值的字符串表示”Prime=nnn“，其中数字的位数明显是变化的；

3、将素数以long类型的二进制表示。

基本策略是，创建一个字节缓冲区，之后创建一系列将三种不同类型数据映射到其中的视图缓冲区。一种简单的方法是每次写入一个素数的数据。

import static java.lang.Math.ceil;
import static java.lang.Math.sqrt;
import static java.lang.Math.min;
import static java.nio.file.StandardOpenOption.*;
import java.nio.file.*;
import java.nio.channels.*;
import java.nio.*;
import java.util.*;
import java.io.IOException;

public class PrimesToFile2 {
  public static void main(String[] args) {
    int primesRequired = 100;                                          // Default count
    if (args.length > 0) {
      try {
        primesRequired = Integer.valueOf(args[0]).intValue();
      } catch (NumberFormatException e) {
        System.out.println("Prime count value invalid. Using default of " + primesRequired);
      }
    }

      long[] primes = new long[primesRequired];                        // Array to store primes
      
      getPrimes(primes);
      Path file = createFilePath("Beginning Java Struff","primes.txt");
      writePrimesFile(primes,file);
    }
      //Calculate enough primes to fill the array
      private static long[] getPrimes(long[] primes) {
          primes[0] = 2L;                                                  // Seed the first prime
          primes[1] = 3L;                                                  // and the second
          // Count of primes found ?up to now, which is also the array index
          int count = 2;
          // Next integer to be tested
          long number = 5L;

          outer:
          for (; count < primes.length; number += 2) {

            // The maximum divisor we need to try is square root of number
            long limit = (long)ceil(sqrt((double)number));

            // Divide by all the primes we have up to limit
            for (int i = 1 ; i < count && primes[i] <= limit ; ++i)
              if (number % primes[i] == 0L)                                // Is it an exact divisor?
                continue outer;                                            // yes, try the next number

            primes[count++] = number;                                      // We got one!
          }
          return primes;
        }
    //Create the path for the named file in the specified directory
    //in the user home directory
    private static Path createFilePath(String directory, String fileName) {
        Path file = Paths.get(System.getProperty("user.home")).resolve(directory).resolve(fileName);
        try {
          Files.createDirectories(file.getParent());                       // Make sure we have the directory
        } catch (IOException e) {
          e.printStackTrace();
          System.exit(1);
        }
        System.out.println("New file is: " + file);
        return file;
    }
    
    //Write the array contents to file
    private static void writePrimesFile(long[] primes, Path file) {
        final int BUFFERSIZE = 100;                                        // Byte buffer size
        try (WritableByteChannel channel = Files.newByteChannel(file, EnumSet.of(WRITE, CREATE))){
          ByteBuffer buf = ByteBuffer.allocate(BUFFERSIZE);
          DoubleBuffer doubleBuf = buf.asDoubleBuffer();
          buf.position(8);
          CharBuffer charBuf = buf.asCharBuffer();
          LongBuffer longBuf = null;
          String primeStr = null;

        for (long prime : primes) {
            primeStr = "prime = " + prime;                                 // Create the string
            doubleBuf.put(0,(double)primeStr.length());                    // Store the string length
            charBuf.put(primeStr);                                         // Store the string
            buf.position(2*charBuf.position() + 8);                        // Position for 3rd buffer
            longBuf = buf.asLongBuffer();                                  // Create the buffer
            longBuf.put(prime);                                            // Store the binary long value
            buf.position(buf.position() + 8);                              // Set position after last value
            buf.flip();                                                    // and flip
            channel.write(buf);                                            // Write the buffer as before

            buf.clear();
            doubleBuf.clear();
            charBuf.clear();
        }
          System.out.println("File written is " + ((FileChannel)channel).size() + " bytes.");
        } catch (IOException e) {
          e.printStackTrace();
        }
    }
}