是否保证模板模板参数调用用户提供的演绎指南 [英] Is it guaranteed that template template parameter invoke user provided deduction guides

问题描述

考虑一个例子:

#include <type_traits>
#include <string>

template <template <class> class TT> //#1
struct Foo {
   static void foo() {
      static_assert(std::is_same_v<decltype(TT("abc")), TT<std::string>>);
   }
};

template <class T>
struct Bar {
    Bar(T) {}
};

template <class T>
Bar(T) -> Bar<std::string>; //#2

int main() {
    Foo<Bar>::foo();
}

[clang] 以及

[clang] as well as [gcc] both seem to use user provided deduction guides (#2) when deducing the template parameter of template template parameter (#1). Is it a standard compliant feature?

推荐答案

是的，这是符合标准的.

Yes, this is standard compliant.

根据 [dcl.type.simple]/2 :

格式为typename _opt 的类型说明符 嵌套名称说明符 _opt template-name 是推断的类类型([dcl.type.class.deduct])的占位符. template-name 应该命名不是注入类名称的类模板.

A type-specifier of the form typename_opt nested-name-specifier_opt template-name is a placeholder for a deduced class type ([dcl.type.class.deduct]). The template-name shall name a class template that is not an injected-class-name.

和 [temp.param]/3 :

其标识符不带省略号的 type-parameter 将其标识符定义为 typedef-name (如果声明时未使用template)或 template -name (如果用template声明)在模板声明的范围内.

A type-parameter whose identifier does not follow an ellipsis defines its identifier to be a typedef-name (if declared without template) or template-name (if declared with template) in the scope of the template declaration.

TT是用template声明的类型参数，这使其成为 template-name ，因此是推导类类型的占位符.所有通常的规则都适用.

TT is a type-parameter declared with template, which makes it a template-name and hence a placeholder for a deduced class type. All the usual rules apply just fine.

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