c++ - 如何抛出文件和行号错误？

Question

我尝试使用BOOST_THROW_EXCEPTION做到这一点，这是example:

#include <boost/throw_exception.hpp>
#include <stdexcept>
void demo_boost_throw()
{
    BOOST_THROW_EXCEPTION(std::runtime_error("boost throw std exception."));
    }
int main() {
    demo_boost_throw();
    return 0;
}

从here中我们可以看到它确实包含文件

#define BOOST_THROW_EXCEPTION(x)\
        ::boost::throw_exception( ::boost::enable_error_info(x) <<\
        ::boost::throw_function(BOOST_THROW_EXCEPTION_CURRENT_FUNCTION) <<\
        ::boost::throw_file(__FILE__) <<\
        ::boost::throw_line((int)__LINE__) )

但是，当我运行该程序时，它不会打印出文件和行。

有current_exception_diagnostic_information()，但这需要捕获并打印。我不想捕获它。我希望e.what()包含throw_function，throw_file和throw_line的额外信息。我怎样才能做到这一点？

Answer 1

如果您有机会访问 C++ 20 ，则可以通过以下方式轻松使用 std::source_location :

#include <string>
#include <sstream>
#include <iostream>
#include <experimental/source_location>

void throw_exception(std::string const& message,
                     std::experimental::source_location const& location = std::experimental::source_location::current()) {
    std::stringstream ss;
    ss << location.file_name()
       << ":"
       << location.line()
       << " "
       << message;
    throw std::runtime_error(ss.str());
}

int main() {
    throw_exception("Random exception");
    return 0;
}

Demo