可以模板化 lambda 函数吗?
在 C++11 中，有没有办法模板化 lambda 函数?或者它天生就太具体而无法模板化?
In C++11, is there a way to template a lambda function? Or is it inherently too specific to be templated?
我知道我可以定义一个经典的模板化类/函子，但问题更像是:该语言是否允许模板化 lambda 函数?
I understand that I can define a classic templated class/functor instead, but the question is more like: does the language allow templating lambda functions?
推荐答案
2018 年更新:C++20 将带有模板化和概念化的 lambdas.该功能已集成到标准草案中.
UPDATE 2018: C++20 will come with templated and conceptualized lambdas. The feature has already been integrated into the standard draft.
2014 年更新:C++14 已于今年发布，现在提供了与本示例中语法相同的多态 lambda.一些主要的编译器已经实现了它.
UPDATE 2014: C++14 has been released this year and now provides Polymorphic lambdas with the same syntax as in this example. Some major compilers already implement it.
目前(在 C++11 中)，遗憾的是没有.多态 lambda 在灵活性和功能方面会非常出色.
At it stands (in C++11), sadly no. Polymorphic lambdas would be excellent in terms of flexibility and power.
它们最终成为单态的最初原因是因为概念.概念使这种代码情况变得困难:
The original reason they ended up being monomorphic was because of concepts. Concepts made this code situation difficult:
template <Constraint T>
void foo(T x)
{
auto bar = [](auto x){}; // imaginary syntax
}
在受约束的模板中，您只能调用其他受约束的模板.(否则无法检查约束.) foo 可以调用 bar(x) 吗?lambda 有什么约束(毕竟它的参数只是一个模板)?
In a constrained template you can only call other constrained templates. (Otherwise the constraints couldn't be checked.) Can foo invoke bar(x)? What constraints does the lambda have (the parameter for it is just a template, after all)?
概念还没有准备好解决这类问题；它需要更多的东西，比如 late_check(在调用之前不检查概念)和其他东西.更简单的方法是放弃所有内容并坚持使用单态 lambda.
Concepts weren't ready to tackle this sort of thing; it'd require more stuff like late_check (where the concept wasn't checked until invoked) and stuff. Simpler was just to drop it all and stick to monomorphic lambdas.
然而，随着 C++0x 中概念的删除，多态 lambda 再次成为一个简单的命题.但是，我找不到任何建议.:(
However, with the removal of concepts from C++0x, polymorphic lambdas become a simple proposition again. However, I can't find any proposals for it. :(
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