为什么编译器尝试实例化一个模板，我不实际在任何地方实例化？ [英] Why does the compiler try to instantiate a template that I don't actually instantiate anywhere?

问题描述

以下更新。

以下是我在main.cpp中的完整代码：

  template< class T> 
 struct other_traits; 
 
 template< class T> 
 struct some_traits {
 typedef decltype（& T :: operator（））Fty; 
 typedef typename other_traits< Fty> :: type type; 
}; 
 
 int main（）{
} 
  

g ++编译正好时Visual Studio 2010出现以下错误：

src\main.cpp（9）：错误C2146：语法错误：缺少';'在标识符'类型之前

--src\main.cpp （10）：参见类模板实例化 some_traits< T> '正在编译

src\main.cpp（9）：error C2868 ：' some_traits< T> :: type '：使用声明的非法语法;预期限定名

（我喜欢最后一个，总wtf。）

我可以把它作为VC10中的一个错误，还是早期实例化有什么好的理由？或者是 decltype 的错误，使编译器认为 Fty 不是从属名称？

---

更新：我试图欺骗编译器，认为 Fty 是使用基类继承的依赖名称：

  template< class T> 
 struct other_traits; 
 
 template< class R，class C> 
 struct other_traits< R（C :: *）（）> {
 typedef R type; 
}; 
 
 template< class Fty> 
 struct base_traits {
 typedef typename other_traits< Fty> :: type type; 
}; 
 
 template< class T> 
 struct some_traits 
：public base_traits< decltype（& T :: operator（））> 
 {};但是编译器仍然尝试在现场实例化/编译一切，抛出这些错误：
 
  src\main.cpp（13）：错误C2039：'type'：不是'other_traits< T>'的成员
 with 
 [
 T = 
] 
 src\main.cpp（19）：参考类模板实例化'base_traits< Fty& $ b with 
 [
 Fty = 
] 
 src\main.cpp（19）：参考类模板实例化'some_traits< T& b src \main.cpp（13）：错误C2146：语法错误：缺少';'在标识符'类型'
 src \main.cpp（13）：错误C4430：缺少类型说明符 -  int假定。注意：C ++不支持default-int 
 src\main.cpp（13）：error C2602：'base_traits< Fty> :: type'不是'base_traits< Fty> 
 with 
 [
 Fty = 
] 
 src\main.cpp（13）：参见'base_traits< Fty> :: type'
 with 
 [
 Fty = 
] 
 src\main.cpp（13）：error C2868：'base_traits< Fty&宣言;预期限定名
与
 [
 Fty = 
] 
  
 
 $ b b 
请注意，模板参数为空。任何想法？
解决方案
似乎是 Bug （如果没有设置特殊标志如以下所说的）。 
以下是 Oracle网站的摘录a> for C ++ templates：
 
 
   7.2.2  b 
 $ b 
 ISO C ++标准允许
开发人员编写模板类
，其中所有成员可能不是
 legal  template 
参数。 只要非法的
成员未实例化，
程序仍然形成良好。 ISO 
 C ++标准库使用这种
技术。但是，
  -template = wholeclass 选项实例化所有成员，因此
不能与这样的模板
类一起使用时用
有问题的模板参数。
 
 
 
Updated below.

The following is the entire code I have in my main.cpp:
template<class T>
struct other_traits;

template<class T>
struct some_traits{
    typedef decltype(&T::operator()) Fty;
    typedef typename other_traits<Fty>::type type;
};

int main(){
}
But I get the following errors with Visual Studio 2010 while g++ compiles just fine:

  
src\main.cpp(9): error C2146: syntax error : missing ';' before identifier 'type'

  --src\main.cpp(10) : see reference to class template instantiation 'some_traits<T>' being compiled

  src\main.cpp(9): error C2868: 'some_traits<T>::type' : illegal syntax for using-declaration; expected qualified-name
(I like that last one, total wtf.)


Can I take that as a bug in VC10 or is there any good reason for the early instantiation? Or is it a bug with decltype that makes the compiler think that Fty is not a dependent name?



Update: I tried to cheat the compiler in thinking that Fty is a dependent name using a base class to inherit from:
template<class T>
struct other_traits;

template<class R, class C>
struct other_traits<R (C::*)()>{
    typedef R type;
};

template<class Fty>
struct base_traits{
    typedef typename other_traits<Fty>::type type;
};

template<class T>
struct some_traits
    : public base_traits<decltype(&T::operator())>
{};
But the compiler still tries to instantiate / compile everything on the spot, spewing these errors:
src\main.cpp(13): error C2039: 'type' : is not a member of 'other_traits<T>'
          with
          [
              T=
          ]
          src\main.cpp(19) : see reference to class template instantiation 'base_traits<Fty>' being compiled
          with
          [
              Fty=
          ]
          src\main.cpp(19) : see reference to class template instantiation 'some_traits<T>' being compiled
src\main.cpp(13): error C2146: syntax error : missing ';' before identifier 'type'
src\main.cpp(13): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
src\main.cpp(13): error C2602: 'base_traits<Fty>::type' is not a member of a base class of 'base_traits<Fty>'
          with
          [
              Fty=
          ]
          src\main.cpp(13) : see declaration of 'base_traits<Fty>::type'
          with
          [
              Fty=
          ]
src\main.cpp(13): error C2868: 'base_traits<Fty>::type' : illegal syntax for using-declaration; expected qualified-name
          with
          [
              Fty=
          ]
Note that the template parameters are empty. Any ideas?
解决方案
It seems to be a Bug (if there is not special flag set as mentioned below).
Following is an excerpt from Oracle website for C++ templates:

  
7.2.2
  
  
The ISO C++ Standard permits
  developers to write template classes
  for which all members may not be
  legal with a given template
  argument. As long as the illegal
  members are not instantiated, the
  program is still well formed. The ISO
  C++ Standard Library uses this
  technique. However, the
  -template=wholeclass option instantiates all members, and hence
  cannot be used with such template
  classes when instantiated with the
  problematic template arguments.


                        
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