重新抛出后丢失的异常信息

Question

重新抛出后，我丢失了异常信息，在上层获得的异常对象就像是全新的。请告知，我如何将信息保留在抛出站点中填充的异常对象中。这是下面的代码

#include <iostream>
#include <stdexcept>

void h()
{
    std::cerr<<"h() throwing exception" << std::endl;
    throw std::runtime_error ("dummy exception for no reason");
    return;
}

void g()
{
    try {
        h();
    }
    catch ( const std::exception &ex)
    {
        std::cerr<<"g() : caught exception : " << ex.what() << std::endl;
        throw ex;
    }
    return;
}

void f()
{
    try {
        g();
    }
    catch ( const std::exception &ex)
    {
        std::cerr<<"f() : caught exception : " << ex.what() << std::endl;
        throw ex;
    }
    return;
}

int main(int ac, char **av)
{
    try {
        f();
    }
    catch ( const std::exception &ex)
    {
        std::cerr<<"main() : caught exception : " << ex.what() << std::endl;
        return -1;
    }
    return 0;
}

输出

h() throwing exception
g() : caught exception : dummy exception for no reason
f() : caught exception : std::exception
main() : caught exception : std::exception