C++ 可变模板模板和完美转发
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Variadic template templates and perfect forwarding
提问by Peter Alexander
This questionon the object generator patterngot me thinking about ways to automate it.
这个关于对象生成器模式的问题让我想到了自动化它的方法。
Essentially, I want to automate the creation of functions like std::make_pair, std::bind1stand std::mem_funso that instead of having to write a different function for each template class type, you could write a single variadic template template function that handles all cases at once. Usage of this function would be like:
从本质上讲，我希望自动像功能的建立std::make_pair，std::bind1st并std::mem_fun使得不必编写为每个模板类类型不同的功能，你可以写一个可变参数模板模板函数来处理所有的情况下一次。此函数的用法如下：
make<std::pair>(1, 2); // equivalent to std::make_pair(1, 2)
make<std::binder2nd>(&foo, 3); // equivalent to std::bind2nd(&foo, 3);
Is it possible to write this function make? I have tried this, but it doesn't work in GCC 4.5 or 4.6:
可以写这个函数make吗？我试过这个，但它在 GCC 4.5 或 4.6 中不起作用：
template <template <typename...> class TemplateClass, typename... Args>
TemplateClass<Args...> make(Args&&... args)
{
return TemplateClass<Args...>(std::forward<Args>(args)...);
}
If I try to call (e.g) make<std::pair>(1, 2)I just get
如果我尝试打电话（例如），make<std::pair>(1, 2)我只会得到
error: no matching function for call to 'make(int, int)'
Have I got the syntax wrong anywhere here?
Or is this right and GCC is wrong?
Or is this just fundamentally impossible in C++0x?
我在这里的任何地方都弄错了语法吗？
或者这是对的，而 GCC 是错的？
或者这在 C++0x 中根本不可能？
[edit]
[编辑]
Proposal N2555seems to suggest that this is allowed and GCC claims to have implemented it in GCC4.4.
提案N2555似乎表明这是允许的，并且GCC 声称已经在 GCC4.4 中实现了它。
采纳答案by Johannes Schaub - litb
That's exactly right. I would expect it to work. So I think that GCC is in error with rejecting that. FWIW:
完全正确。我希望它起作用。所以我认为 GCC 拒绝这样做是错误的。FWIW：
#include <utility>
template <template <typename...> class TemplateClass, typename... Args>
TemplateClass<Args...> make(Args&&... args)
{
return TemplateClass<Args...>(std::forward<Args>(args)...);
}
int main() {
make<std::pair>(1, 2);
}
// [js@HOST2 cpp]$ clang++ -std=c++0x main1.cpp
// [js@HOST2 cpp]$
回答by Luc Danton
This is probably a GCC quirk. I can get the following to work with a dev snapshot (I don't have a copy of 4.6 right now):
这可能是 GCC 的怪癖。我可以获得以下内容来处理开发快照（我现在没有 4.6 的副本）：
template<
template<typename...> class TemplateClass
, typename... Args
, typename Result = TemplateClass<Args...>
// Also works with the arguably more correct
// , typename Result = TemplateClass<
// typename std::decay<Args>::type...
// >
>
Result
make(Args&&... args)
{ /* as before */ }
回答by Puppy
This is quite wrong- take make_shared, for example. The point of make_sharedis that there are run-time efficiency savings for using it. But what would happen if I tried to use make<std::shared_ptr>? Don't think that would quite work out. Or how about types where only some of the constructor arguments are template arguments, and the rest aren't? For example, make<std::vector, int>(other_vector.begin(), other_vector.end());- the types of the iterators don't participate, but you pass them in anyway.
这是非常错误的——make_shared举个例子。关键make_shared是使用它可以节省运行时效率。但是如果我尝试使用会发生什么make<std::shared_ptr>？不要认为那会很奏效。或者只有一些构造函数参数是模板参数而其余不是的类型呢？例如，make<std::vector, int>(other_vector.begin(), other_vector.end());- 迭代器的类型不参与，但无论如何你都传递它们。
It's impossible to write a generic makefunction.
写一个泛型make函数是不可能的。
As for the Standard, well, it could easily have been removed since then. You'd have to check the FDIS.
至于标准，嗯，从那时起它很容易被删除。你必须检查 FDIS。
