C ++ catch枚举值作为异常 [英] C++ catch enum value as exception

问题描述

我尝试使用已定义其例外的外部 c / c ++ 库：

 枚举MY_ERRORS {
 ERR_NONE = 0，
 ERR_T1，
 ERR_T2，
}然后在代码中抛出异常如下：
  if（...）{
 throw ERR_T1; 
  
作为C ++中的编程的新手，我会做一些事情：
  try {
 call_to_external_library（）; 
} catch（??? err）{
 printf（发生错误：％s\\\
，err）; 
} catch（...）{
 printf（发生意外异常。\\\
）; 
} 
  
 如何确定引发的内容？ / p> 
 
解决方案
您需要编写代码来处理catch块中的枚举类型：
  try {
 call_to_external_library（）; 
} catch（MY_ERRORS err）{//< ------------------------ HERE 
 printf（An error occurred ：％s\\\
，err）; 
} catch（...）{
 printf（发生意外异常。\\\
）; 
} 
  
 
I am trying to use an external C++ library which have defined its exceptions as:
enum MY_ERRORS {
    ERR_NONE = 0,
    ERR_T1,
    ERR_T2,
};
Then in the code exceptions are thrown like this:
if(...) {
    throw ERR_T1;
Being new to programming in C++, I would do something like:
try {
    call_to_external_library();
} catch(??? err) {
    printf("An error occurred: %s\n", err);
} catch(...) {
    printf("An unexpected exception occurred.\n");
}
How do I determine what was thrown?
解决方案
You will need to write your code to handle the type of enumeration in the catch block:
try {
    call_to_external_library();
} catch(MY_ERRORS err) {      // <------------------------ HERE
    printf("An error occurred: %s\n", err);
} catch(...) {
    printf("An unexpected exception occurred.\n");
}


                        
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