澄清人工计算的环复杂性 [英] Clarifying the manual count of Cyclomatic Complexity

问题描述

让我们假设我们有这样的代码:

Let's assume that we have a code like this:

switch(y)
{
case 1: case 2: case 3:
    function();
    break;
case 4: case 5: case 6:
    function_2();
    break;
}

我们可以在这里将CC值设为6 + 1吗?为什么要添加值1?如果将CC值视为7，那么这是独立路径的数量吗?

Can we get the CC value as 6+1 here? Why a value of 1 is added? If the CC value is considered as 7, is that the number of independent paths?

如果上面考虑了失败的情况怎么办?可能只有两条唯一的路径，2 +1 = 3

What if a fall through scenario is considered above? As only possible two unique paths are there, 2 +1 =3

以上哪个是正确的，还是两个都是正确的?

Which of the above are correct or are the both of them correct?

推荐答案

我们知道，CC = P+1.

在这里，P =谓词节点(条件)的数量= 2

Here, P = number of predicate nodes (conditions) = 2

条件数将为2，因为:

Case分支可以覆盖多个替代值或范围，例如 案例1、2、5至10.由于他们没有引入其他分支来决定 ，它们也不会增加循环复杂度.

Case branch can cover several alternative values or ranges, such as Case 1, 2, 5 To 10. As they introduce no additional branches to decide on, they do not increase cyclomatic complexity either.

来源:此处

所以，CC = 2 + 1 = 3

So, CC = 2+1 = 3

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