Timeline for Cyclomatic complexity with two IFs - why it is 3?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2013 at 13:30 | comment | added | John V | Well, because I cannot understand why it is not 4 - doesn't it miss one path? I'm sure there is a mathematical proof why it doesn't but in the examle, I would say it does. | |
| Jan 17, 2013 at 13:26 | comment | added | Kristof Claes | Ah, I see. But what's the problem then? The article says it's 3, you say it has to be 3 according to the CC formula, yet you ask why it isn't 4? | |
| Jan 17, 2013 at 13:22 | comment | added | John V | I dont think so as the formulae for CC is: Number of IFs - Number of end points + 2. For this code it makes 3 (2-1+2), regardless of what the statements are. | |
| Jan 17, 2013 at 13:19 | comment | added | Kristof Claes | If the operations change, it is possible to have four instead of three possible outcomes. In this specific scenario, only three outcomes are possible. The reason is because the second operation is the opposite of the first one. So both happening is the same as none happening. | |
| Jan 17, 2013 at 13:17 | comment | added | John V | What do you mean? In this example yes, but different statements are run. Imagine there is a=a+5 and the other one is a=1000;. | |
| Jan 17, 2013 at 13:12 | history | answered | Kristof Claes | CC BY-SA 3.0 |