我尝试保存一些整数并以二进制格式将其加倍到txt文件。我认为数字可以正确保存,但是问题是当我尝试使用readInt()读取整数时。例如,当整数值为12时,该整数的程序输出为208797696,并且具有更多整数的程序始终以EOF异常结束。 readDouble()没问题,它可以正常工作。
import java.util.Scanner;
import java.io.*;
public class CandyFruit {
public static void main(String[] args) throws IOException {
int i;
int int_sc ;
double double_sc;
String string_sc;
int seek_w = 0; //seek for writing
String map = "";
Scanner scan = new Scanner(System.in);
try (RandomAccessFile write = new RandomAccessFile("fgh.txt", "rw")) {
for (;;) {
System.out.println("1 - write int, 2 - write double, 0 - break");
i = scan.nextInt();
if (i == 0) break;
System.out.println("Your choice:");
if (i == 1) {
int_sc = scan.nextInt();
write_int(write, int_sc, seek_w);
map = map + "1";
seek_w += 4; //seek growth for int
}
else if (i == 2) {
double_sc = scan.nextDouble();
write_double(write, double_sc, seek_w);
map = map + "2";
seek_w += 8; //seek growth for double
}
}
}
//catch
try (RandomAccessFile read = new RandomAccessFile("fgh.txt", "rw")) {
int seek_r = 0;
for (int o = 0; o < map.length(); o++) {
char c = map.charAt(o);
if (c == '1') {
read_int(read, seek_r);
seek_r += 4; //seek growth for int
}
else if (c == '2') {
read_double(read, seek_r);
seek_r += 8; //seek growth for double
}
}
}
//catch
}
public static void write_int (RandomAccessFile write, int scanned, int s) throws IOException {
write.seek(s);
write.write(scanned);
}
public static void write_double(RandomAccessFile write, double scanned, int s) throws IOException {
write.seek(s);
write.writeDouble(scanned);
}
public static void read_int(RandomAccessFile read, int s) throws IOException {
read.seek(s);
System.out.println("Int is: "+read.readInt());
}
public static void read_double(RandomAccessFile read, int s) throws IOException {
read.seek(s);
System.out.println("Double is: "+read.readDouble());
}
}
答案 0 :(得分:1)
您的write_int
方法使用RandomAccessFile.write()
进行写入,该方法将 byte 写入文件。使用readInt()
时,您会读取该单字节和三字节的其他任意数据(因为您没有写这些数据)。
您应该在RandomAccessFile.writeInt()
中使用write_int
。