SonarQube 暗示 var 不为空

Question

在下面的示例中，SonarQube 抱怨model.toString()isnot null且 ( model == null) 始终为false，需要一些帮助来了解可以采取哪些措施来修复它。因为书签在 if 语句中被初始化为变量，并且显然将是null.

public static class Mapper implements DataStore.ModelMapper<Membership, 
MembershipPassDTO> {            
    @Override
    public MembershipPassDTO mapModel(Membership model) {
        VALogger.e("SULOD Membership Mapper", "" + model.toString());

        if (model == null) {
            return new MembershipPassDTO(model, "", "", "", "", "", "");
        }
        return new MembershipPassDTO(model, model.getVitalityMembershipId(), model.getMembershipNumber(), model.getCustomerNumber(),
                model.getVitalityStatus(), model.getMembershipStartDate(), model.getMembershipStatus());
    }
}

Answer 1

每当你调用model.toString()并且你的模型确实为 null时，你的方法就会抛出一个NullPoiunterException.

但如果model.toString()没有抛出 aNullPointerEcxeption那么很明显，那model == null总是假的......

嗯 - 这就是 SonarQube 想要告诉你的......

如果你想摆脱警告，你可以按照jensgram 的建议进行操作：

VALogger.e("SULOD Membership Mapper", ""+model); //implicitly calls toString()