问题描述
我有一些不同成员的类，所有类都有以下类型的操作。
I have a number of classes with differing members, all of which have operations of the following type
::basedata::Maindata maindata;
::basedata::Subdata subinfo("This goes into the subinfo vector");
subinfo.contexts(contextInfo);
maindata.subdata().push_back(subinfo);
请注意，我要求如何设置通用模板来执行这些操作。我不能为每种类型的maindata和subinfo设置一个特殊情况。我还需要能够看到如何从我的主代码调用模板。我已经能够设置一个模板如果 maindata.subdata（）存在，但是如果不存在，调用模板时仍然会收到编译失败。这是创建形式的模板
Note that I am asking how to set up generalized templates to perform these actions. I cannot set up a special case for each type of maindata and subinfo. I also need to be able to see how to call the template from my main code. I have been able to set up a template if maindata.subdata() exists, but keep getting a compilation failure on a call to the template if it does not exist. That is create the template of the form
perform_Push(maindata.subdata(), subinfo);
，因此无论maindata.subdata（）是否存在，都可以编译。
so that it can be compiled whether or not maindata.subdata() exists or not.
我可以接受构建的模板，以便主代码可以显示
I could accept templates that build so that the main code can show
bool retval
retval = hasmember(maindata, subdata);
if (retval)
{
buildmember(maindata.subdata, subinfo);
setAttributes(subinfo, data);
perform_Push(maindata.subdata(), subinfo)
}
else
{
// Perform alternate processing
}
到目前为止，当调用的模板应该是void时，if中的代码不会被编译。
As of now, the code inside the if would not compile when the templates being called should just be void.
当:: basedata :: Maindata总是被定义，:: basedata :: Subdata可能或可能不被定义，这取决于我的代码正在构建的库的版本。子数据定义为属于maindata的向量，因此定义了push_back（）操作。在任何情况下，有太多类型的subData为每个类型创建一个单独的模板作为模板中的T :: Subdata。
While ::basedata::Maindata is always defined, ::basedata::Subdata may or may not be defined depending on the release of libraries that my code is being build with. subdata is defined as a vector belonging to maindata which therefore has the push_back() operation defined. There are too many types of subData to create a separate template for each type as T::Subdata within a template in any case.
也就是说，如果subdata是唯一的case，我可以创建模板T的专业化为:: maindata :: subdata和一个通用模板T。
That is, if subdata was the only case, I could create a specialization of the template T as ::maindata::subdata and a generic Template T.
我没有任何控制包含文件或库，为此，我不能创建一个#define的变量来测试与预编译器。有没有一个好的方法设置一个模板，将允许这个工作？我可以使用返回布尔值true（成功）或false（没有这样的定义）的模板，并在运行时调用替代处理。我不需要有一个替代模板。
I do not have any control of the include files or library that for this so that I cannot create a #define of a variable to test with the pre-compiler. Is there a good way of setting up a template that would allow this to work? I could use a template that returns a boolean true (success) or false (no such definition) and call the alternate processing at run time. I would not need to have an alternate template.
基本上，我问如何应用SFINAE这种特殊情况。
Basically, I am asking how to apply SFINAE to this particular situation.
我已经设法找出了我需要做什么来设置基本模板
I have managed to figure out what I need to do to set up the basic template
如果我有最基本的操作
maindata.subdata().push_back(data)
b $ b
我可以定义一个形式的模板，
I can define a template of the form,
<template class T, typename D>
auto doPush(D data) -> decltype(T.pushback(data), void())
{
T.push_back(data);
}
，呼叫
doPush<maindata.subdata()>(data);
但是，问题是如何在maindata还没有成员子数据时设置它。
However, the problem would be how to set it up when maindata does not yet have a member subdata.
推荐答案
您可以使用此模板获取一个布尔值，告诉您是否存在成员类型 Subdata 在通用类型 T 中。只有当 T 是一个结构/类而不是一个命名空间时，它才能工作。
You can use this templates to obtain a boolean value that tell you if exist member type Subdata in a generic type T. This works only if T is a struct/class not a namespace.
#include <type_traits>
template <class T, class V = void>
struct hasSubdata
{
enum { value = false };
};
template <class T>
struct hasSubdata<T, typename std::enable_if< std::is_same<typename T::Subdata, typename T::Subdata>::value >::type>
{
enum { value = true };
};
struct basedata1
{
struct Subdata {};
};
struct basedata2
{
};
#include <iostream>
int main ()
{
std::cout << "basedata1: " << hasSubdata<basedata1>::value << std::endl;
std::cout << "basedata2: " << hasSubdata<basedata2>::value << std::endl;
}
但是你不能使用普通的if，因为编译器检查正确性所有的可能性。
你必须以类似的方式（很丑陋）：
But you can't use a normal if because the compiler checks the correctness of all the possibilities.You have to act in a similar way (pretty ugly):
template <class T, bool = hasSubdata<T>::value>
struct SubdataUser
{
static void foo ()
{
std::cout << "I can use SubData member :)" << std::endl;
typename T::Subdata subinfo ();
}
};
template <class T>
struct SubdataUser<T, false>
{
static void foo ()
{
std::cout << "I can not :(" << std::endl;
}
};
int main ()
{
SubdataUser<basedata1>::foo ();
return 0;
}
据我所知，你不能有一个模板 hasMember< Type，Member> :: value 因为如果 Member
Unfortunately to my knowledge, you can not have a template hasMember<Type,Member>::value because if Member does not exist, compilation fails.
But you might like a solution of this type $ c> #include< type_traits>
#include< iostream>
struct basedata1
{
struct Subdata1 {};
struct Subdata2 { };
struct Subdata3 {};
};
struct basedata2
{
struct Subdata1 {};
// struct Subdata2 { };
struct Subdata3 {};
};
template< class ...>
struct Require
{
枚举{value = true};
};
template< class T，bool = true>
struct Impl
{
static void foo
{
std :: cout<< 至少有一个所需的成员不可用:(<<< std :: endl;
}
};
模板< class T>
struct Impl< T，Require< typename T :: Subdata1，
typename T :: Subdata2，
typename T :: Subdata3> :: value>
{
static void foo（）
{
std :: cout<<所有成员都可用:)< std :: endl;
typename T :: Subdata2 my_var;
}
};
int main（int argc，char * argv []）
{
Impl< basedata1> :: foo
Impl< basedata2> :: foo（）;
return 0;
}
#include <type_traits>#include <iostream>struct basedata1{ struct Subdata1 {}; struct Subdata2 {}; struct Subdata3 {};};struct basedata2{ struct Subdata1 {}; //struct Subdata2 {}; struct Subdata3 {};};template <class...> struct Require { enum { value = true }; };template <class T, bool = true> struct Impl { static void foo () { std::cout << "At least one of the members required is not available :(" << std::endl; } };template <class T> struct Impl<T, Require< typename T::Subdata1, typename T::Subdata2, typename T::Subdata3 >::value > { static void foo () { std::cout << "All members are available :)" << std::endl; typename T::Subdata2 my_var; } };int main( int argc, char* argv[] ){ Impl<basedata1>::foo (); Impl<basedata2>::foo (); return 0;}
我希望这有助于
这篇关于变量类型声明的C ++模板的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！