构造函数中引发的异常的处理程序是什么？ [复制]

我是 C++ 异常处理的新手。我心中的规则是，

1. 如果没有找到异常处理程序调用链（函数调用的堆栈），然后终止函数被调用。
2. 处理程序是一个catch {} block.

但是，我无法理解以下行为。

#include <iostream>
#include <exception>
using namespace std;

struct X {
  X() try { throw exception(); }
  catch (exception &e) {
    cout << "Exception caught in constructor!" << endl;
  }
};

int main() {

  try {
    throw exception();
  }
  catch (exception &e) {
    cout << "Exception caught in function." << endl;
  }
  cout << "After Exception being caught in function" << endl;

  try {
    X x;
  }
  catch (exception &e) {
    cout << "Why exception is caught again!" << endl;
  }

  return 0;
}

输出是

Exception caught in function. After Exception being caught in function Exception caught in constructor! Why exception is caught again!

问题一：看起来 X 的构造函数中抛出的异常被捕获（或处理）了两次。或者为什么构造函数后面的 catch{} 块不被算作构造函数中异常的处理程序？

如果我不放X x; in the try{}阻止并抓住它main()，输出为：

Exception caught in function. After Exception being caught in function Exception caught in constructor! terminate called after throwing an instance of 'std::exception' what(): std::exception Aborted (core dumped)

问题2：当我们没有时调用默认的终止（）函数X x;在 try 块中？

With:

X() try { throw exception(); }
catch (exception &e) {
    cout << "Exception caught in constructor!" << endl;
}

X对象尚未完全构造，因此 catch 不能忽略异常，而应重新抛出异常。

IMO，更清楚什么时候X有一个成员抛出...（X() try : member(0) {} catch(exception&) {} ).

请注意，您可以使用normal在构造函数块内尝试 catch：

X() {
    try { throw exception(); }
    catch (exception &e) {
        cout << "Exception caught in constructor!" << endl;
    }
}

这更自然。