Question

所以我有一个基本的加密器，在解密后，我想将该文件中的数字字符串转换为int。

public class Encryption {

//Basic Encryptor/Decryptor
public Encryption() throws IOException {
    StringBuffer num = new StringBuffer("100");
    encrypt(num);

    File f = new File("C:\\fun\\a");
    if(f.mkdirs()) {
    File f2 = new File(f,"b.txt");
    FileWriter fw = new FileWriter(f2);
    BufferedWriter bw = new BufferedWriter(fw); 
    bw.write(num.toString());
    bw.close();

    Scanner s = new Scanner(f2);
    String aaa = new String(s.nextLine());
    StringBuffer bbb = new StringBuffer(aaa);
    decrypt(bbb);
    int b = Integer.parseInt(aaa.toString());

    System.out.println(b);



    }

}

//ENCRYPTOR
public static void encrypt(StringBuffer word) {
    for(int i = 0; i < word.length(); i++) {
        int temp = 0;
        temp = (int)word.charAt(i);
        temp = temp * 9;
        word.setCharAt(i, (char)temp);
    }
}

//DECRYPTOR
public static void decrypt(StringBuffer word) {
    for(int i = 0; i < word.length(); i++) {
        int temp = 0;
        temp = (int)word.charAt(i);
        temp = temp / 9;
        word.setCharAt(i, (char)temp);
    }
}


public static void main(String[] args) {    
    try {
        @SuppressWarnings("unused")
        Encryption e = new Encryption();
    } catch (IOException e) {
        e.printStackTrace();
    }
}
  }

我试图将字符串缓冲区转换为字符串然后转换为int。当我检查文件时，它仍然是加密的，并且控制台有解密错误。

Exception in thread "main" java.lang.NumberFormatException: For input string: "???"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at Encryptor.Encryption.<init>(Encryption.java:28)
at Encryptor.Encryption.main(Encryption.java:62)

哪个指向int b = Integer.parseInt(aaa.toString());。

Answer 1

我认为您唯一的问题是您正在尝试解析aaa StringBuffer的内容。 aaa StringBuffer仍包含加密内容。您正在将bbb缓冲区传递给decrypt方法，因此这是您应该尝试解析为int的缓冲区：

int b = Integer.parseInt(bbb.toString());

将StringBuffer加密转换为int

1 个答案: