相关疑难解决方法(0)

g ++和gcc有什么区别？哪些应该用于一般的c ++开发？

所有,我在我的mac os x 10.8中编写了这样的代码,当我使用"gcc use_new.cpp -o use_new"来编译它时,它会抛出错误的消息,如下所示:

Undefined symbols for architecture x86_64:
  "std::basic_ostream<char, std::char_traits<char> >::operator<<(std::basic_ostream<char, std::char_traits<char> >& (*)(std::basic_ostream<char, std::char_traits<char> >&))", referenced from:
      _main in ccr2vrRQ.o
  "std::basic_ostream<char, std::char_traits<char> >::operator<<(void const*)", referenced from:
      _main in ccr2vrRQ.o
  "std::basic_ostream<char, std::char_traits<char> >::operator<<(double)", referenced from:
      _main in ccr2vrRQ.o
  "std::basic_ostream<char, std::char_traits<char> >::operator<<(int)", referenced from:
      _main in ccr2vrRQ.o
  "std::basic_ostream<char, std::char_traits<char> >::operator<<(unsigned long)", referenced from:
      _main in ccr2vrRQ.o
  "std::ios_base::Init::Init()", referenced from:
      __static_initialization_and_destruction_0(int, int) in ccr2vrRQ.o
  "std::ios_base::Init::~Init()", referenced from:
      __static_initialization_and_destruction_0(int, int) in ccr2vrRQ.o
  "std::cout", referenced from:
      _main in ccr2vrRQ.o
  "std::basic_ostream<char, …

是否可以在专用模板类中访问非类型模板参数的值？

如果我有专门化的模板类:

   template <int major, int minor> struct A {
       void f() { cout << major << endl; }
   }

   template <> struct A<4,0> {
       void f() { cout << ??? << endl; }
   }

我知道上面的情况,硬编码值4和0而不是使用变量很简单,但我有一个更大的类,我是专门的,我希望能够访问值.

在A <4,0>中是否可以访问major和minor值(4和0)？或者我必须在模板实例化时将它们分配为常量:

   template <> struct A<4,0> {
       static const int major = 4;
       static const int minor = 0;
       ...
   }