c++ - 通过引用捕获异常对象、临时对象、生命周期问题

Question

考虑以下代码:

#include <iostream>
#include <stdexcept>

void foo()
{
    throw std::runtime_error("How long do I live?");
}

int main()
{
    try
    {
        foo();
    }
    catch (std::runtime_error& e)
    {
        std::cout << e.what() << std::endl;
    }
}

为什么我可以通过引用捕获异常，std::runtime_error("How long do I live?") 不是右值吗？

为什么异常对象在 catch block 中仍然存在？

抛出的异常对象到底存放在哪里？他们的生命周期是多少？

Answer 1

在 C++ 标准中，第 15.1.4 段:

The memory for the temporary copy of the exception being thrown is allocated in an unspecified way, except as noted in 3.7.3.1. The temporary persists as long as there is a handler being executed for that exception. In particular, if a handler exits by executing a throw; statement, that passes control to another handler for the same exception, so the temporary remains. When the last handler being executed for the exception exits by any means other than throw; the temporary object is destroyed and the implementation may deallocate the memory for the temporary object; any such deallocation is done in an unspecified way. The destruction occurs immediately after the destruction of the object declared in the exception-declaration in the handler.

请注意，在 C++ 标准谈话中，处理程序表示具有正确参数类型的 catch block 。