代碼分析-JAVA 求大神幫看下這段代碼 怎麼修改 救命啊

JAVA 求大神幫看下這段代碼 怎麼修改 救命啊

真心看不懂

public static char[] a(char[] paramArrayOfChar) {

    char[] arrayOfChar1 = new char[12];
    char[] arrayOfChar2 = new char[12];
    arrayOfChar2[0] = '«';
    arrayOfChar2[1] = ' ';
    arrayOfChar2[2] = 'o';
    arrayOfChar2[3] = '/';
    arrayOfChar2[4] = '\037';
    arrayOfChar2[5] = '\036';
    arrayOfChar2[6] = 'š';
    arrayOfChar2[7] = 'E';
    int i = 0;
    int j = 0;
    int k;

    if ((paramArrayOfChar[j] < '0') || (paramArrayOfChar[j] > '9'))
        return null;

    while (true) {

        if (j >= 15)
            return null;

        do {
            arrayOfChar1[i] = (char) (arrayOfChar1[i] ^ arrayOfChar2[i]);
            arrayOfChar1[8] = 'W';
            arrayOfChar1[9] = 'Û';
            arrayOfChar1[11] = '\000';
            arrayOfChar1[10] = '\000';
            k = 0;
            if (k < 10)
                break;              
            arrayOfChar1[10] = (char) (0xFF & arrayOfChar1[10]);
            arrayOfChar1[11] = (char) (0xFF & arrayOfChar1[11]);
            return arrayOfChar1;



            arrayOfChar1[i] = (char) ('￐' + paramArrayOfChar[j]);
        } while (j >= 14);
        if ((paramArrayOfChar[(j + 1)] < '0')
                || (paramArrayOfChar[(j + 1)] > '9'))
            return null;
        arrayOfChar1[i] = (char) (arrayOfChar1[i] + ('￐' + paramArrayOfChar[(j + 1)] << 4));
        arrayOfChar1[i] = (char) (arrayOfChar1[i] ^ arrayOfChar2[i]);
        j = 1 + (j + 1);
        i++;
    }

    int m = k & 0x1;
    Boolean.valueOf(false);
    Boolean localBoolean;
    if (m == 0) {
        localBoolean = Boolean.valueOf(false);
        label267: if (!localBoolean.booleanValue())
            break label304;
        arrayOfChar1[11] = (char) (arrayOfChar1[11] + arrayOfChar1[k]);
    }
    while (true) {
        k++;
        break;
        localBoolean = Boolean.valueOf(true);
        break label267;
        label304: arrayOfChar1[10] = (char) (arrayOfChar1[10] + arrayOfChar1[k]);
    }
}

最佳回答：

反編譯的代碼麼？看著很怪，而且那些看上去亂碼的字符似乎沒有反編譯對。