?故障过程
1、上午的时候,QA同学突然说,测试自动化的流程突然过不去了,问我是不是最近对线上做了某些修改。当时第一反应是不可能
2、通过QA同学提供的test case,在测试环境通过curl发送请求,发现果然广告返回值跟预期不符。
3、通过git log对比,发现近期只有一个switch语句有修改。
4、尝试在代码中加入log语句,发现日志输出果然跟QA的错误结果一致,至此原因找到。
故障原因
下面是错误代码
switch?(dsp_res->bid_type())?{ ????????case?0: ????????{ ??????????auto?info?=?dsp_response->add_dsp_res_infos(); ??????????info->set_dsp_id(item.dsp_id()); ??????????if?(dsp_res->has_cache_duration())?{ ????????????info->set_duration(dsp_res->cache_duration()); ??????????} ??????????if?(dsp_res->has_quality())?{ ????????????info->set_ratio(dsp_res->quality()); ??????????} ????//?do?sth ????????} ????????case?1: ??????????break; ????????case?2: ????????{ ??????????auto?info?=?dsp_response->add_dsp_res_infos(); ??????????if?(dsp_res->has_cache_duration())?{ ????????????info->set_duration(dsp_res->cache_duration()); ??????????} ??????????if?(dsp_res->has_quality())?{ ????????????info->set_ratio(dsp_res->quality()); ??????????} ??????????info->set_dsp_id(item.dsp_id()); ??????????std::unordered_set<std::string>?adids; ??????????for?(auto?elem?:?dsp_res->ad_ids())?{ ????????????adids.insert(elem); ??????????} ????//?do?sth ?????????? ????????} ????????case?3:?//?此case为新增 ????????{ ??????????auto?info?=?dsp_response->add_dsp_res_infos(); ??????????info->set_dsp_id(item.dsp_id()); ??????????if?(dsp_res->has_cache_duration())?{ ????????????info->set_duration(dsp_res->cache_duration()); ??????????} ??????????if?(dsp_res->has_quality())?{ ????????????info->set_ratio(dsp_res->quality()); ??????????} ????//?do?sth ????????} ????????default: ??????????break; ??????}
发现,当dsp_res->bid_type() == 2的时候,也会执行 case 3的部分,然后尝试在上面各个"do sth" 后面,加上break,结果符合预期,bug搞定。
深思
为什么在未增加新case之前,test case能通过呢?仔细找QA问了下case的逻辑,原来,case每次都会返回bid_type = 2。此处,我们再贴一次之前的代码:
switch?(dsp_res->bid_type())?{ ????????case?0: ????????{ ??????????auto?info?=?dsp_response->add_dsp_res_infos(); ??????????info->set_dsp_id(item.dsp_id()); ??????????if?(dsp_res->has_cache_duration())?{ ????????????info->set_duration(dsp_res->cache_duration()); ??????????} ??????????if?(dsp_res->has_quality())?{ ????????????info->set_ratio(dsp_res->quality()); ??????????} ????//?do?sth ????????} ????????case?1: ??????????break; ????????case?2: ????????{ ??????????auto?info?=?dsp_response->add_dsp_res_infos(); ??????????if?(dsp_res->has_cache_duration())?{ ????????????info->set_duration(dsp_res->cache_duration()); ??????????} ??????????if?(dsp_res->has_quality())?{ ????????????info->set_ratio(dsp_res->quality()); ??????????} ??????????info->set_dsp_id(item.dsp_id()); ??????????std::unordered_set<std::string>?adids; ??????????for?(auto?elem?:?dsp_res->ad_ids())?{ ????????????adids.insert(elem); ??????????} ????//?do?sth ????????} ????????default: ??????????break; ??????}
由于switch每次都会进入case 2的子逻辑,该逻辑后面就是default,然后break,没问题。但是增加了新case 3之后,因为case 2 和 case 3后面都没有break,导致会把case 2 和 case 3的代码都执行了,直到退出循环或者遇到break。此处列下switch case的三个规则:switch...case的三个规则:
- 既无成功匹配,又无default子句,那么swtich语句块什么也不做;
- 无成功匹配,但有default,那么swtich语句块做default语句块的事;
- 有成功匹配,没有break,那么成功匹配后,一直执行,直到遇到break。
看来我们的线上bug是因为遇到了第三个规则导致。
扩展
语句体中不包含break
#include?<stdio.h> int?main() { ????int?iChoice?=?0; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",&iChoice); ????switch?(iChoice) ????{ ????case?1: ????????printf("case?1?!\n"); ????case?2: ????????printf("case?2?!\n"); ????case?3: ????????printf("case?3?!\n"); ????default: ????????printf("default?!\n"?); ????} ????return?0; }
当输入choice 为 1的时候
当输入choice 为 2的时候
原因:
在上面的示例中,如果iChoice等于1,则执行主体的所有语句,因为在开关主体中没有出现break语句。如果ichoice等于2,则由于没有break语句,因此控制跳至情况2并执行以下所有情况。
一个执行语句被多个case命中
void?TestFunction(void) { ????printf("Demo?code\n"); } int?main() { ????int?iChoice?=?0; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",?&iChoice); ????switch?(?iChoice?) ????{ ????case?1: ????case?2: ????case?3: ????????//Calling?function ????????TestFunction(); ????????break; ????case?4: ????????printf("Wow?Test?paas?!"); ????????break; ????default: ????????printf("default?!\n"?); ????} ????return?0; }
输出为
原因:
对于case 1 2 3,都会执行到TestFunction
存在一样的case标签
#include?<stdio.h> int?main() { ????int?iChoice???=?0; ????int?i?=?0; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",?&iChoice); ????switch?(?iChoice?) ????{ ????case?1: ????????i++; ????????break; ????case?3: ????????i?=?i?+?2; ????????break; ????case?3: ????????i?=?i?+?3; ????????break; ????default: ????????printf("default?!\n"?); ????????break; ????} ????printf("Value?of?i?=?%d",i); ????return?0; }
输出
原因
case标签不能重复,否则编译器不能确定进入哪个标签
case值为浮点数
#include?<stdio.h> int?main() { ????int?iChoice???=?0; ????int?i?=?0; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",?&iChoice); ????switch?(iChoice) ????{ ????case?1: ????????i++; ????????break; ????case?2.5: ????????i?=?i?+?2; ????????break; ????case?3: ????????i?=?i?+?3; ????????break; ????default: ????????printf("default?!\n"?); ????} ????printf("Value?of?i?=?%d",i); ????return?0; }
输出:
原因:
switch 中的参数必须可以转换成一个整数
将default 语句放在正文的其他地方
#include?<stdio.h> int?main() { ????int?iChoice??=?0; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",?&iChoice); ????switch?(iChoice) ????{ ????default: ????????printf("Bad?Input?!\n"); ????????break; ????case?1: ????????printf("Your?enter?choice?is?1\n"); ????????break; ????case?2: ????????printf("Your?enter?choice?is?2\n"); ????????break; ????case?3: ????????printf("Your?enter?choice?is?3\n"); ????????break; ????} ????return?0; }
输出:
case标签值必须为常量
#include?<stdio.h> int?main() { ????int?iChoice??=?0; ????int?Label?=?1; ????printf("Enter?your?choice?=?"); ????scanf(?"%d",?&iChoice); ????switch?(iChoice) ????{ ????case?Label: ????????printf("Your?enter?choice?is?1\n"); ????????break; ????case?2: ????????printf("Your?enter?choice?is?2\n"); ????????break; ????case?3: ????????printf("Your?enter?choice?is?3\n"); ????????break; ????default: ????????printf("Bad?Input?!\n"); ????????break; ????} ????return?0; }
输出:
嵌套开关
#include?<stdio.h> void?nestedSwitchDemo(int?input1,?int?input2) { ????switch?(input1) ????{ ????case?1: ????????printf("Your?enter?choice?is?1\n"); ????????switch?(input2?) ????????{ ????????case?1: ????????????printf("Enter?Sub?choice?is?1\n"); ????????????break; ????????case?2: ????????????printf("Enter?Sub?choice?is?2\n"); ????????????break; ????????} ????????break; ????case?2: ????????printf("Your?enter?choice?is?2\n"); ????????break; ????case?3: ????????printf("Your?enter?choice?is?3\n"); ????????break; ????default: ????????printf("Bad?Input?!\n"); ????????break; ????} } int?main() { ????int?iChoice??=?1; ????int?iSubChoice?=?1; ????//Calling?function ????nestedSwitchDemo(iChoice,iSubChoice); ????return?0; }
输出:
心得
平时编码中,一定要注意编码规范,每个case都写好对应的break,不要学习这种骚操作,稍不注意就可能出现线上故障。