java SONAR 抱怨要改变条件，使其不总是评估为“假”

SONAR complaining to change the condition so that it does not always evaluate to "false"

提问by Awa

public String generateURLSafeToken(String username, char[] password) throws CredentialTokenException {
    this.tokenValid = false;

    String token = null;
    if ((username.length() < 1) || (username == null)) {
        throw new CredentialTokenException("Username cannot be an empty string or null.");
    }
    if ((password.length < 1) || (password == null)) {
        throw new CredentialTokenException("Password cannot be an empty or null.");
    }

I am facing this error in line 4 and line 7 (username == null and password == null)

我在第 4 行和第 7 行中遇到此错误（用户名 == 空和密码 == 空）

And I need this part in my code. I am trying isEmpty() instead of null but facing problems in that also . What is an alternate way or the solution to fix this SONAR error

我的代码中需要这部分。我正在尝试 isEmpty() 而不是 null 但也面临着问题。修复此 SONAR 错误的替代方法或解决方案是什么

回答by Tunaki

The conditions which always evaluates to falseare username == nulland password == null.

总是评估为的条件false是username == null和password == null。

Let's take the example of username. The operator ||is short-circuitingmeaning it won't evaluate the right hand side if the left hand side is true. Basically, there are 2 cases:

让我们以 为例username。运算符||是短路的，这意味着如果左侧是 ，则不会评估右侧true。基本上有2种情况：

• The usernamegiven is not null. The condition username.length() < 1is evaluated
  • If the result is true, we return directly and enter the ifbranch
  • If the result is false, we try to evaluate username == null. But since the usernamegiven is not null, this alwaysevaluate to false.
• The usernamegiven is null. The condition username.length() < 1is evaluated. This actually stops right there: it will throw a NullPointerExceptionand will not evaluate the right hand side.

• 在username给出的不是null。条件username.length() < 1被评估
  • 如果结果是true，我们直接返回并进入if分支
  • 如果结果是false，我们尝试评估username == null。但由于username给定的不是null，因此始终评估为false。
• 在username给出的null。username.length() < 1评估条件。这实际上就停在那里：它会抛出 aNullPointerException并且不会评估右手边。

Therefore, you can see that whenever the username == nullcondition was actually evaluated, the result was always false. This is what the SonarQube warning is telling you.

因此，您可以看到，无论何时username == null实际评估条件，结果始终为false。这就是 SonarQube 警告告诉你的。

The solution here is to reverse your 2 conditions. Consider having

这里的解决方案是反转您的 2 个条件。考虑拥有

if (username == null || username.length() < 1)

instead. If you start over and go through each case, you'll notice that none of the expressions will always have the same result:

反而。如果您重新开始并检查每种情况，您会注意到没有一个表达式总是具有相同的结果：

• The usernamegiven is not null. First condition clearly evaluates to falseand the second is evaluated, which may return trueor false.
• The usernamegiven is null. The first condition clearly evaluated to trueand short-circuits.

• 在username给出的不是null。第一个条件明确评估为false，第二个条件评估，可能返回true或false。
• 在username给出的null。第一个条件明确评估为true短路。