如果我试图在C++中的catch块中抛出一些内容,为什么它会导致终止？

Question

我有以下C++代码,它给了我一个惊喜.问题是如果我抛出一些东西,除了在catch块内重新抛出,程序将通过调用abort终止并在GCC4中给出错误消息,"在抛出'int'实例后调用终止".如果我只是用"扔"; 重新扔进catch区,一切都会好的.

#include <iostream> 
#include <exception>
#include <stdexcept>

using namespace std;

int main()
{
    try{
        throw std::string("first throw");
    }
    catch(std::string &x){
        try{
            std::cout << x << std::endl;
//          throw;  // if I use this line, all is fine.
            throw int(2); // but if I use this line, it causes Abort() to be called
        }
        catch (int &k){
            throw;
        }
        catch(...)
        {
            cout << "all handled here!"<< endl;
        }
    }
    catch(...){
        std::cout<< "never printed" << endl;
    }
}

Answer 1

如果你扔了int,那么就不会被处理; 它会被内部catch (int &k)处理程序捕获,它会重新抛出它; 并且没有外部处理程序来捕获重新抛出的异常,因为您已经在外部catch块中.所以在这种情况下,terminate由于未处理的异常而被调用.

如果你重新抛出string,那么它被内部catch(...)处理程序捕获; 这不会重新抛出,因此异常已被处理.

Answer 2

你throw不在任何try处理程序内,因此它会abort被调用.

这是你的代码与缩进清理了一些和内联的一些注释:

#include <iostream> 
#include <exception>
#include <stdexcept>

using namespace std;

int main()
{
    try {
        throw std::string("first throw");
    }
    catch (std::string &x) {
        try {
            std::cout << x << std::endl;
            // throw;  // if I use this line, all is fine.
            throw int(2); // but if I use this line, it causes Abort() to be called
        }
        catch (int &k) {
            // Catches and rethrows exception. Not inside a try thus leads to abort.
            throw;
        }
        catch (...) {
            // Will handle the case where the string is rethrown instead. No abort.
            cout << "all handled here!"<< endl;
        }
    }
    catch (...) {
        std::cout<< "never printed" << endl;
    }
}