template-name<TT> 是推导上下文吗？
Is template-name<TT> a deduced context?
[temp.deduct.type] paragraph 8 列出了所有推断的上下文，但它似乎不包括 template-name<TT> where template-name 指的是 class 模板，TT 指的是模板模板参数。这是推导上下文吗？
如果是，为什么？
如果不是，请考虑以下代码：
template<template<typename> class U, template<typename> class V>
struct foo {};
template<template<typename> class U>
struct foo<U, U> {};
int main() {}
此代码编译 under Clang 7.0.0 and GCC 8.0.1，这意味着编译器认为部分特化比主模板更特化，这意味着主模板中的 U 和 V 被成功推导反对 foo<U, U>。这是编译器错误吗？
没错，template-name<TT>是不是推导的上下文。这与这里无关。 Plain TT 是 一个推断的上下文，这就是你在这里
non-type 模板参数 I 的推导上下文 template-name<I> 意味着当参数 Foo<5> 参数 [=] 时可以推导 5 16=]。
template-name<TT> 不是推导上下文的原因很简单。如果 template-name<TT> 是合法的，那么 template-name 就必须采用模板模板参数。这意味着它本身必须是模板模板模板 ("TTT")。语言定义中只有这么多递归。
[编辑]
在您的示例中，您正在为 U 推导 V 以查看它是否是专业化。两者都是模板模板。你不是想推断是否 Foo<U> isFoo. Therefore your deduced context isTT, nottemplate-name.`。
要回答评论的第 2 点，TT 是一个推导的上下文，因为该形式已明确列出。 template-name<TT> 的意思是，当 TT 用作已知模板的参数时，TT 必须是可推导的。但是什么类型的模板接受模板模板参数呢？即假设的TTT.
这段有很多问题，包括你指出的那个。 Core issue 2328 有一个不错的列表：
The presentation style of 17.9.2.5 [temp.deduct.type] paragraph 8
results in a specification that is unclear, needlessly verbose, and
incomplete. Specific problems include:
What does it mean for P and A to have one of a set of forms? Do they both have to have that form? (That doesn't happen; typically,
only P contains template parameters)
In the introductory sentence, aren't T, TT, and i supposed to be the names of template parameters rather than template arguments?
In T[i], it appears we can deduce i, but not T (T can only be deduced in the form T[<i>integer-constant</i>])
What is an integer-constant supposed to be?
What is a cv-list?
Why can we not deduce const T from T? (Apparently you only get to deduce if both or neither type have a cv-list, whatever a
cv-list is.)
We have extreme redundancy because, for instance, there is no way to say “in T (T::*)(T), you can deduce any of those Ts, and it's OK
if some of the positions don't have a T”. So we have seven (!) forms
of that construct, for all cases except the one where none of the
three positions contain a T.
We have special case rules for pointers to member functions, even though they're not a special case and should be covered by the rule
for pointers to members and the rule for functions.
We do not allow deducing a template template parameter's value from a template template argument — there is a TT<T> form, a TT<i>
form, a <i>template-name</i><T> form, and a
<i>template-name</i><i> form, but no TT<TT> form
nor <i>template-name</i><TT> form.
看起来编辑设法摆脱了 cv-list，至少，自从提交问题以来。现在只是 cv。 （cv-list 有点可笑的错误，因为 [语法] 说 -list 后缀用于 comma-separated 列表...)
[temp.deduct.type] paragraph 8 列出了所有推断的上下文，但它似乎不包括 template-name<TT> where template-name 指的是 class 模板，TT 指的是模板模板参数。这是推导上下文吗？
如果是，为什么？
如果不是，请考虑以下代码：
template<template<typename> class U, template<typename> class V>
struct foo {};
template<template<typename> class U>
struct foo<U, U> {};
int main() {}
此代码编译 under Clang 7.0.0 and GCC 8.0.1，这意味着编译器认为部分特化比主模板更特化，这意味着主模板中的 U 和 V 被成功推导反对 foo<U, U>。这是编译器错误吗？
没错，template-name<TT>是不是推导的上下文。这与这里无关。 Plain TT 是 一个推断的上下文，这就是你在这里
non-type 模板参数 I 的推导上下文 template-name<I> 意味着当参数 Foo<5> 参数 [=] 时可以推导 5 16=]。
template-name<TT> 不是推导上下文的原因很简单。如果 template-name<TT> 是合法的，那么 template-name 就必须采用模板模板参数。这意味着它本身必须是模板模板模板 ("TTT")。语言定义中只有这么多递归。
[编辑]
在您的示例中，您正在为 U 推导 V 以查看它是否是专业化。两者都是模板模板。你不是想推断是否 Foo<U> isFoo. Therefore your deduced context isTT, nottemplate-name.`。
要回答评论的第 2 点，TT 是一个推导的上下文，因为该形式已明确列出。 template-name<TT> 的意思是，当 TT 用作已知模板的参数时，TT 必须是可推导的。但是什么类型的模板接受模板模板参数呢？即假设的TTT.
这段有很多问题，包括你指出的那个。 Core issue 2328 有一个不错的列表：
The presentation style of 17.9.2.5 [temp.deduct.type] paragraph 8 results in a specification that is unclear, needlessly verbose, and incomplete. Specific problems include:
What does it mean for
PandAto have one of a set of forms? Do they both have to have that form? (That doesn't happen; typically, onlyPcontains template parameters)In the introductory sentence, aren't
T,TT, andisupposed to be the names of template parameters rather than template arguments?In
T[i], it appears we can deducei, but notT(Tcan only be deduced in the formT[<i>integer-constant</i>])What is an
integer-constantsupposed to be?What is a
cv-list?Why can we not deduce
const TfromT? (Apparently you only get to deduce if both or neither type have acv-list, whatever acv-listis.)We have extreme redundancy because, for instance, there is no way to say “in
T (T::*)(T), you can deduce any of thoseTs, and it's OK if some of the positions don't have aT”. So we have seven (!) forms of that construct, for all cases except the one where none of the three positions contain aT.We have special case rules for pointers to member functions, even though they're not a special case and should be covered by the rule for pointers to members and the rule for functions.
We do not allow deducing a template template parameter's value from a template template argument — there is a
TT<T>form, aTT<i>form, a<i>template-name</i><T>form, and a<i>template-name</i><i>form, but noTT<TT>form nor<i>template-name</i><TT>form.
看起来编辑设法摆脱了 cv-list，至少，自从提交问题以来。现在只是 cv。 （cv-list 有点可笑的错误，因为 [语法] 说 -list 后缀用于 comma-separated 列表...)