C语言 在 switch 语句中使用枚举类型
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Using enum type in a switch statement
提问by tomocafe
I am using a switch statement to return from my main function early if some special case is detected. The special cases are encoded using an enum type, as shown below.
如果检测到某些特殊情况，我将使用 switch 语句尽早从主函数返回。特殊情况使用枚举类型进行编码，如下所示。
typedef enum {
NEG_INF,
ZERO,
POS_INF,
NOT_SPECIAL
} extrema;
int main(){
// ...
extrema check = POS_INF;
switch(check){
NEG_INF: printf("neg inf"); return 1;
ZERO: printf("zero"); return 2;
POS_INF: printf("pos inf"); return 3;
default: printf("not special"); break;
}
// ...
return 0;
}
Strangely enough, when I run this, the string not specialis printed to the console and the rest of the main function carries on with execution.
奇怪的是，当我运行它时，字符串not special被打印到控制台，而主函数的其余部分继续执行。
How can I get the switch statement to function properly here? Thanks!
我怎样才能让 switch 语句在这里正常运行？谢谢！
回答by ldav1s
No caselabels. You've got gotolabels now. Try:
没有case标签。你goto现在有标签了。尝试：
switch(check){
case NEG_INF: printf("neg inf"); return 1;
case ZERO: printf("zero"); return 2;
case POS_INF: printf("pos inf"); return 3;
default: printf("not special"); break;
}
回答by Jonathan Leffler
You're missing the all-important case:
你错过了最重要的case：
switch(check){
case NEG_INF: printf("neg inf"); return 1;
case ZERO: printf("zero"); return 2;
case POS_INF: printf("pos inf"); return 3;
default: printf("not special"); break;
}
You created some (unused) labels with the same names as your enumeration constants (which is why it compiled).
您创建了一些与枚举常量同名的（未使用的）标签（这就是编译的原因）。
回答by Deepu
You haven't used the keyword "case". The version given below will work fine.
您还没有使用关键字“case”。下面给出的版本将正常工作。
typedef enum {
NEG_INF,
ZERO,
POS_INF,
NOT_SPECIAL
} extrema;
int main(){
extrema check = POS_INF;
switch(check){
case NEG_INF: printf("neg inf"); return 1;
case ZERO: printf("zero"); return 2;
case POS_INF: printf("pos inf"); return 3;
default: printf("not special"); break;
}
return 0;
}
