凯撒加密。数组在73中的索引73中越界

Question

我正在为一个数学演示的凯撒加密代码工作。今天，我一直在为我的演示文稿编码。早上这个代码已经开始工作了。但是现在在学校，我遇到了一个例外，我认为，的数超过了我的符号表的长度。

例外情况：

如果有人能查看我的代码，我会很高兴，因为我，以我3个月的编码经验，不能理解发生了什么。

//Alphanumeric table
    final static char[] symbs = new char[] {'.',',','1','2','3','4','5','6','7','8','9','0','/','z','y','x','w','v','u','t','s','r','q','p','o','n','m','l','k','j','i','h','g','f','e','d','c','b','a',' '
                    ,'ß','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','ä','ö','ü','Ü','Ö','Ä'};
    final static int charlength = 73;

    //var global key
    int key = 0;

    //en- and decryption vars
    char[] letters = new char[255];
    char[] storg = new char[255];
    char[] sterg = new char[255];
    StringBuilder str = new StringBuilder();
    int cnt = 0;
    int cnt2 = 0;
    int c = 0;
    String bakToStr;

textField1.addKeyListener(new KeyAdapter() {
            @Override
            public void keyTyped(KeyEvent e) {
                super.keyTyped(e);
                char iNumber = e.getKeyChar();
                if (!(Character.isDigit(iNumber))
                        || iNumber == KeyEvent.VK_BACK_SPACE
                        || iNumber == KeyEvent.VK_DELETE) {
                    e.consume();
                    JOptionPane.showMessageDialog(contentPane, "Nur Zahlen werden als schlüssel akzeptiert", "Achtung!", JOptionPane.INFORMATION_MESSAGE);
                }
            }
        });

public String encrypt(String Text) {
        letters = new char[255];
        storg = new char[255];
        sterg = new char[255];
        str = new StringBuilder();
        cnt = 0;
        cnt2 = 1;
        c = 0;
        bakToStr = "";
        for (int i = 0; i <Text.length(); i++) {
            letters[i] = Text.charAt(i);
        }
        for (; cnt < Text.length(); cnt++) {
            for (;letters[cnt] != symbs[cnt2];cnt2++) {
            }
            c = (cnt2 + key) %charlength;
            storg[cnt] = symbs[c];
            cnt2 = 0;
        }
        str.append(storg);
        bakToStr = String.valueOf(str);
        return bakToStr;
    }

    public String decrypt(String Text) {
        letters = new char[255];
        storg = new char[255];
        sterg = new char[255];
        str = new StringBuilder();
        cnt = 0;
        cnt2 = 1;
        c = 0;
        bakToStr = "";
        for (int i = 0; i <Text.length(); i++) {
            letters[i] = Text.charAt(i);
        }
        for (; cnt < Text.length(); cnt++) {
            for (;letters[cnt] != symbs[cnt2];cnt2++) {
            }
            c = (cnt2 - key) %charlength;
            while (c <= 0) {
                c += charlength;
            }
            storg[cnt] = symbs[c];
            cnt2 = 0;
        }
        str.append(storg);
        bakToStr = String.valueOf(str);
        return bakToStr;
    }




Answer 1

请考虑这句话：

while (c <= 0) {
    c += charlength;
}

如果c==0会发生什么？我认为它将跳转到73，即charlengt。因此，这里更好的方法是使条件while(c<0)而不是while(c<=0)。

但是，我认为最主要的问题是第三个for循环：

for (;letters[cnt] != symbs[cnt2];cnt2++)

我希望它能有所帮助：）