使用atexit（）时不为本地对象调用析构函数

Question

class Static {
public:
    ~Static() 
        {
        std::cout << "Static dtor\n";
        }
    };
class Sample {
public:
    ~Sample() 
        {
        std::cout << "Sample dtor\n";
        }
    };

class Local {
public:
    ~Local() 
        {
        std::cout << "Local dtor\n";
        }
    };

Static static_variable; // dtor of this object *will* be called
void atexit_handler()
    {
    std::cout << "atexit handler\n";
    }
int main()
    {
    Local local_variable; 
    const int result = std::atexit(atexit_handler); 
    Sample static_variable; // dtor of this object *will not* be called
    std::exit(EXIT_SUCCESS);//succesful exit
    return 0;
    }

Answer 1

调用析构函数不是关于atexit，而是关于exit。

我一般不认为std::exit有什么好的C++编程。实际上，这和std::atexit

extern "C"   int atexit( void (*func)() ); // in <cstdlib>

来自C标准库。在查看您的示例时，我相信您已经看到了http://en.cppreference.com/w/cpp/utility/program/exit，您还看到了

Static static_variable; 

int mainactual()
{
    Local local_variable; 
    Sample static_variable;
    throw std::exception("EXIT_FAILURE"); // MS specific
    // ... more code
}
int main()
{
    try 
    { 
        mainactual() 
    }catch ( std::exception & e )
    {
        return EXIT_FAILURE;
    }
    return EXIT_SUCCESS;
}