具有多个出口点的代码中的循环复杂度 [英] Cyclomatic Complexity in piece of code with multiple exit points

问题描述

我有这种验证密码的方法：

I have this method that validates a password:

/**
 * Checks if the given password is valid.
 * 
 * @param password The password to validate.
 * @return {@code true} if the password is valid, {@code false} otherwise.
 */
public static boolean validatePassword(String password) {
    int len = password.length();
    if (len < 8 || len > 20)
        return false;
    boolean hasLetters = false;
    boolean hasDigits = false;
    for (int i=0; i<len; i++) {
        if (!Character.isLetterOrDigit(password.charAt(i)))
            return false;
        hasDigits = hasDigits || Character.isDigit(password.charAt(i));
        hasLetters = hasLetters || Character.isLetter(password.charAt(i));
    }
    return hasDigits && hasLetters;
}

让我们关注圈复杂度数：它的值是什么？

Let's focus on the cyclomatic complexity number: what is its value?

指标1.3.6 表示现在是7，但我不能找到七个独立的路径：我只有五个！而且维基百科并没有多大帮助-我想怎么使用这个公式π-s + 2 ？

Metrics 1.3.6 says it's 7, but I cannot really find seven independent paths: I only find 5! And Wikipedia didn't help out much—how am I suppose to use this formula π - s + 2?

如果有，我有2个， 1个 和3个出口点，但我被卡住了：我必须计算入口点吗？我应该对第一个 if 进行两次计数，因为它有两个条件吗？

I have 2 if's, 1 for and 3 exit points but I'm stuck: do I have to count the entry point? Should I count twice the first if since it has two conditions?

编辑：

好，现在我发现Cyclomatic Number是7。这意味着有7条独立的路径，因此如果我愿意，我应该能够找到7个不同的测试用例覆盖100％的代码，对吗？

Ok, now I found out that Cyclomatic Number is 7. This means that there are 7 independent paths and so I should be able to find 7 different test cases if I would to cover 100% of code, am I right?

嗯，我仍然找不到最后一个！
我找到了这些：

Well, I still cannot found the last one! I found these:

1. 有效：asdf1234


2. 太短：asdf123


3. 太长：asdfsgihzasweruihioruldhgobaihgfuiosbhrbgtadfhsdrhuorhguozr


4. 无效字符：asdf * 123


5. 全数字：12345678


6. 无数字：asdfghjk


7. wtf ???

1. Valid: asdf1234
2. Too short: asdf123
3. Too long: asdfsgihzasweruihioruldhgobaihgfuiosbhrbgtadfhsdrhuorhguozr
4. Invalid character: asdf*123
5. All-digit: 12345678
6. No-digits: asdfghjk
7. wtf???

推荐答案

我认为，诀窍在于计算逻辑运算符。

I think the trick is that the logical operators are counted.

基于您的Metrics链接（ http://metrics.sourceforge.net/ ）在 McCabe圈复杂度部分下：

Based off of your Metrics link (http://metrics.sourceforge.net/) under the McCabe Cyclomatic Complexity section:

1初始流

3个决策点（if，for，if）

3 decision points (if,for,if)

3个条件逻辑运算符（||，|| ,, ||）

3 conditional logic operators (||,||,||)

总计：7

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