gcc 可以编译可变参数模板，而 clang 不能
我正在阅读一些名为 C++11 和 C++14 概述，由 Leor Zolman 先生提出.在第 35 页，他介绍了一种使用 decltype 进行求和运算的方法.
I'm reading some slides named An Overview of C++11 and C++14 presented by Mr. Leor Zolman. At Page 35 he introduces a way to do the sum operation with decltype.
struct Sum {
template <typename T>
static T sum(T n) {
return n;
}
template <typename T, typename... Args>
/// static T sum(T n, Args... rest) {
static auto sum(T n, Args... rest) -> decltype(n + sum(rest...)) {
return n + sum(rest...);
}
};
当为Sum::sum(1, 2.3, 4, 5); 使用这个片段时，clang-3.6(from svn) 无法用 -std=c++ 编译它11/-std=c++1y 但 gcc-4.9 成功了.当然，如果没有对返回类型进行类型推导，两者都可以编译，但这涉及类型转换，无法得到预期的结果.
When using this snippets forSum::sum(1, 2.3, 4, 5); clang-3.6(from svn) fails to compile this with -std=c++11/-std=c++1y but gcc-4.9 succeeds. Of course without type deduction for the return type both compile, but that involves type conversion and cannot get the expected result.
那么这是否表示一个 clang 错误，或者是因为 gcc 扩展(就 c++11 或 c++14 而言)?
So does this indicate a clang bug, or is because of a gcc extension(in respect of c++11 or c++14)?
推荐答案
Clang 的行为是正确的.这是一个 GCC 错误(并且演示文稿中的声明也不正确).§3.3.2 [basic.scope.pdecl]/p1,6:
Clang's behavior is correct. This is a GCC bug (and the claim in the presentation is also incorrect). §3.3.2 [basic.scope.pdecl]/p1,6:
1 名称的声明点紧随其后完整的声明符(第 8 条)及其初始化程序(如果有)之前，除非如下所述.
1 The point of declaration for a name is immediately after its complete declarator (Clause 8) and before its initializer (if any), except as noted below.
6 类成员声明点后，成员名可以在其类的范围内查找.
6 After the point of declaration of a class member, the member name can be looked up in the scope of its class.
第 3.3.7 节 [basic.scope.class]/p1 说
And §3.3.7 [basic.scope.class]/p1 says
以下规则描述了在类中声明的名称的范围.
The following rules describe the scope of names declared in classes.
1) 在类中声明的名称的潜在范围不仅包括名称声明点之后的声明区域，还有所有的函数体，默认参数，exception-specifications 和 brace-or-equal-initializers该类中的非静态数据成员(包括嵌套类).
1) The potential scope of a name declared in a class consists not only of the declarative region following the name’s point of declaration, but also of all function bodies, default arguments, exception-specifications, and brace-or-equal-initializers of non-static data members in that class (including such things in nested classes).
trailing-return-types 不在该列表中.
尾随返回类型是声明符的一部分 (§8 [dcl.decl]/p4):
The trailing return type is part of the declarator (§8 [dcl.decl]/p4):
declarator:
ptr-declarator
noptr-declarator parameters-and-qualifiers trailing-return-type
因此 sum 的可变参数版本不在其自己的trailing-return-type 范围内，并且无法通过名称查找找到.
and so the variadic version of sum isn't in scope within its own trailing-return-type and cannot be found by name lookup.
在 C++14 中，只需使用实际返回类型推导(并省略尾随返回类型).在 C++11 中，你可以使用一个类模板和一个简单转发的函数模板:
In C++14, simply use actual return type deduction (and omit the trailing return type). In C++11, you may use a class template instead and a function template that simply forwards:
template<class T, class... Args>
struct Sum {
static auto sum(T n, Args... rest) -> decltype(n + Sum<Args...>::sum(rest...)) {
return n + Sum<Args...>::sum(rest...);
}
};
template<class T>
struct Sum<T>{
static T sum(T n) { return n; }
};
template<class T, class... Args>
auto sum(T n, Args... rest) -> decltype(Sum<T, Args...>::sum(n, rest...)){
return Sum<T, Args...>::sum(n, rest...);
}
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