C ++ 14 lambda的默认参数类型推导取决于先前的参数 [英] C++14 lambda's default argument type deduction depending on preceding arguments
问题描述
这不是C ++ 14版本吗?
Is this not valid as C++14?
auto f = [](auto x, auto y = std::decay_t<decltype(x)>{}) { };
f(0);
我期望它大致等于
auto f = [](int x, int y) { };
f(0, int{});
GCC 6.3和Clang 4.0都不接受我的代码.
Neither GCC 6.3 nor Clang 4.0 accepted my code.
- http://ideone.com/b7b4SK GCC
- http://ideone.com/EyLYaL Clang
这与我对C ++模板推导阶段缺乏了解有关吗?长达1400页的规格实际上对我的问题有明确的答案吗?
Is it related to my lack of understanding of C++ template deduction phases? Does the 1400 pages long spec actually has an explicit answer to my question?
总而言之，我的问题实际上可以简化为这段代码(没有lambda，单个参数)，并且在C ++ 14下无效(感谢@BaummitAugen和@NirFriedman)
To summarize, my problem can in fact be reduced to this piece of code (free of lambda, single parameter) and it is invalid under C++14 (thanks @BaummitAugen and @NirFriedman)
template <typename T>
void f(T x = 0) { }
int main() {
f();
}
推荐答案
编译器正确拒绝了您的代码，这实际上不是有效的C ++ 14.
The compilers are correct to reject your code, it is indeed not valid C++14.
在标准(此处使用N4141)中，
In the standard (using N4141 here) we have
对于通用lambda，闭包类型具有公共内联函数调用 操作员成员模板(14.5.2)，其template-parameter-list由一种发明的类型template- 在lambda的parameter-declaration-clause中，按出现顺序为每次出现auto设置参数.
For a generic lambda, the closure type has a public inline function call operator member template (14.5.2) whose template-parameter-list consists of one invented type template- parameter for each occurrence of auto in the lambda’s parameter-declaration-clause, in order of appearance.
(5.1.2/4 [expr.prim.lambda]).因此，您的通话等同于拨打某些电话
(5.1.2/4 [expr.prim.lambda]). So your call is equivalent to a call to some
template <class T1, class T2>
auto operator() (T1 x, T2 y = std::decay_t<decltype(x)>{});
现在
如果仅在非推导中使用模板参数 上下文且未明确指定，模板参数推导失败.
If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.
(14.8.2/4 [temp.deduct.type])和
(14.8.2/4 [temp.deduct.type]) and
非推论上下文是:
[...]
-具有默认参数的函数参数的参数类型中使用的模板参数 用于进行参数推导的调用中.
The non-deduced contexts are:
[...]
- A template parameter used in the parameter type of a function parameter that has a default argument that is being used in the call for which argument deduction is being done.
(14.8.2/5 [temp.deduct.type])使您的通话格式不正确.
(14.8.2/5 [temp.deduct.type]) makes your call ill-formed.
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