圈复杂度 - 绘制此 java 语句的控制流图

Question

有人可以帮忙吗？

while (x > level)
    x = x – 1;
x = 0

Answer 1

可以使用提供的公式计算圈复杂度 here。

Cyclomatic complexity = E - N + P 
where,
  E = number of edges in the flow graph.
  N = number of nodes in the flow graph.
  P = number of nodes that have exit points

对于您的情况，图表应如下所示：

---------------                ----------
|  x > level  |----- NO ------>| x = x-1|
|-------------|                ----|-----
       |      |---------------------
       |
      Yes
       |
-------|----------
| End while (if) |
-------|----------
       |
       |
   ---------
   |  x = 0 |
   ----------

（非ASCII艺术人）

所以，cyclomatic complexity 应该是：

E = 4, N = 4, P = 2 => Complexity = 4 - 4 + 2 = 2

[编辑] Ira Baxter 很好地指出了如何为 Java、C#、C++ 等语言简化此计算。但是，必须仔细执行识别条件，如图所示 here:

- Start with a count of one for the method.
- Add one for each of the following flow-related elements that are found in the method.
    Returns - Each return that isn't the last statement of a method.
    Selection - if, else, case, default.
    Loops - for, while, do-while, break, and continue.
    Operators - &&, ||, ?, and :
    Exceptions - catch, finally, throw, or throws clause.

圈复杂度 - 绘制此 java 语句的控制流图

Cyclomatic complexity - draw control flow graph for this java statement

testing

metrics

control-flow-graph