如何确定类是否有特定的模板成员函数？ [英] How to determine whether a class has a particular templated member function?

问题描述

我想知道是否可以扩展SFINAE方法来检测类是否具有某个成员函数（如下所述：

I was wondering if it's possible to extend the SFINAE approach to detecting whether a class has a certain member function (as discussed here:

是否有一个技术C ++知道一个类是否有给定签名的成员函数？
http://stackoverflow.com/questions/87372/is-there-a-technique-in-c-

"Is there a Technique in C++ to know if a class has a member function of a given signature?" http://stackoverflow.com/questions/87372/is-there-a-technique-in-c-to-know-if-a-class-has-a-member-function-of-a-given-s

），以支持模板化成员函数（to-template-a-class-has-a-member-function-of-a-given- ？例如。以便能够检测以下类中的函数foo：

) to support templated member functions? E.g. to be able to detect the function foo in the following class:

struct some_class {
   template < int _n > void foo() { }
};

我认为这可能是为了foo的特定实例化，如果 void foo< 5>（）是成员）：

I thought it might be possible to do this for a particular instantiation of foo, (e.g. check to see if void foo< 5 >() is a member) as follows:

template < typename _class, int _n >
class foo_int_checker {

  template < typename _t, void (_t::*)() >
  struct sfinae { };

  template < typename _t >
  static big
  test( sfinae< _t, &_t::foo< _n > > * );

  template < typename _t >
  static small
  test( ... );

public:

  enum { value = sizeof( test< _class >( 0 ) ) == sizeof( big ) };

};

然后执行 foo_int_checker< some_class，5> :: value 以检查 some_class 是否具有成员 void foo< 5>（）。然而，在MSVC ++ 2008上，这总是返回 false ，而g ++在 test（sfinae <_t，& _t :: foo& _n>>）;

Then do foo_int_checker< some_class, 5 >::value to check whether some_class has the member void foo< 5 >(). However on MSVC++ 2008 this always returns false while g++ gives the following syntax errors at the line test( sfinae< _t, &_t::foo< _n > > );

test.cpp:24: error: missing `>' to terminate the template argument list
test.cpp:24: error: template argument 2 is invalid
test.cpp:24: error: expected unqualified-id before '<' token
test.cpp:24: error: expected `,' or `...' before '<' token
test.cpp:24: error: ISO C++ forbids declaration of `parameter' with no type

这两个都似乎失败，因为我试图从一个类型本身获取模板函数实例化的地址模板参数。有没有人知道这是可能的，还是因为某些原因而被标准禁止？

Both seem to fail because I'm trying to get the address of a template function instantiation from a type that is itself a template parameter. Does anyone know whether this is possible or if it's disallowed by the standard for some reason?

编辑：看起来我错过了 template 语法来使g ++正确地编译上面的代码。如果我改变的位，我得到的函数的地址为& _t :: template foo<然后程序编译，但我得到与MSVC ++相同的行为（ value 始终设置为 false ）

It seems that I missed out the ::template syntax to get g++ to compile the above code correctly. If I change the bit where I get the address of the function to &_t::template foo< _n > then the program compiles, but I get the same behaviour as MSVC++ (value is always set to false).

如果我注释掉 ... c> test 强制编译器选择另一个，我在g ++中得到以下编译器错误：

If I comment out the ... overload of test to force the compiler to pick the other one, I get the following compiler error in g++:

test.cpp: In instantiation of `foo_int_checker<A, 5>':
test.cpp:40:   instantiated from here
test.cpp:32: error: invalid use of undefined type `class foo_int_checker<A, 5>'
test.cpp:17: error: declaration of `class foo_int_checker<A, 5>'
test.cpp:32: error: enumerator value for `value' not integer constant

其中第32行是枚举{value = sizeof（test< _class>（0））== sizeof（big）}; 线。不幸的是，这似乎并没有帮助我诊断问题:(。MSVC ++给出类似的不明确的错误：

where line 32 is the enum { value = sizeof( test< _class >( 0 ) ) == sizeof( big ) }; line. Unfortunately this doesn't seem to help me diagnose the problem :(. MSVC++ gives a similar nondescript error:

error C2770: invalid explicit template argument(s) for 'clarity::meta::big checker<_checked_type>::test(checker<_checked_type>::sfinae<_t,&_t::template foo<5>> *)'

在同一行。

如果我从一个特定的类而不是一个模板参数获取地址（即，而不是& _t :: template foo <_n> 我做& some_class :: template foo< _n> ）然后我得到正确的结果，但是我的检查器类被限制为检查单个类（ some_class

What's strange is that if I get the address from a specific class and not a template parameter (i.e. rather than &_t::template foo< _n > I do &some_class::template foo< _n >) then I get the correct result, but then my checker class is limited to checking a single class (some_class) for the function. Also, if I do the following:

template < typename _t, void (_t::*_f)() >
void
f0() { }

template < typename _t >
void
f1() {
  f0< _t, &_t::template foo< 5 > >();
}

并调用 f1< some_class>（）然后我不会得到一个编译错误& _t :: template foo< 5> 。这表明问题只出现在当在SFINAE上下文中从本身是模板参数的类型获得模板成员函数的地址时。 Argon！

and call f1< some_class >() then I DON'T get a compile error on &_t::template foo< 5 >. This suggests that the problem only arises when getting the address of a templated member function from a type that is itself a template parameter while in a SFINAE context. Argh!

推荐答案

在Boost.MPL中已经实现了类似的东西，它被称为BOOST_MPL_HAS_XXX_TRAIT_DEF。请参阅：

There is something similar already implemented in Boost.MPL, it is called "BOOST_MPL_HAS_XXX_TRAIT_DEF". See:

http://www.boost.org/doc/libs/1_41_0/libs/mpl/doc/refmanual/has-xxx-trait-def.html

它可以检测类是否具有给定的类型。

It can detect if the class have a given named type.

对于你的具体情况，而不是传递函数指针作为参数（void（_t :: *）（）），尝试使用它在你的方法的主体，即像：

Also, for your specific case, instead of passing the function pointer as a parameter (void (_t::*)()), try to use it in the body of your method, i.e., something like:

template < typename _t >
static big test( sfinae<_t> )
{
  &_t::foo<_n>;
}

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