問題描述

我正在閱讀一些名為 C++11 和 C++14 概述，由 Leor Zolman 先生提出.在第 35 頁(yè)，他介紹了一種使用 decltype 進(jìn)行求和運(yùn)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...);
  }
};

當(dāng)為Sum::sum(1, 2.3, 4, 5); 使用這個(gè)片段時(shí)，clang-3.6(from svn) 無法用 -std=c++ 編譯它11/-std=c++1y 但 gcc-4.9 成功了.當(dāng)然，如果沒有對(duì)返回類型進(jìn)行類型推導(dǎo)，兩者都可以編譯，但這涉及類型轉(zhuǎn)換，無法得到預(yù)期的結(jié)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.

那么這是否表示一個(gè) clang 錯(cuò)誤，或者是因?yàn)?gcc 擴(kuò)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)?

推薦答案

Clang 的行為是正確的.這是一個(gè) GCC 錯(cuò)誤(并且演示文稿中的聲明也不正確).§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 名稱的聲明點(diǎn)緊隨其后完整的聲明符(第 8 條)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 類成員聲明點(diǎn)后，成員名可以在其類的范圍內(nèi)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 節(jié) [basic.scope.class]/p1 說

And §3.3.7 [basic.scope.class]/p1 says

以下規(guī)則描述了在類中聲明的名稱的范圍.

The following rules describe the scope of names declared in classes.

1) 在類中聲明的名稱的潛在范圍不僅包括名稱聲明點(diǎn)之后的聲明區(qū)域，還有所有的函數(shù)體，默認(rèn)參數(shù)，exception-specifications 和 brace-or-equal-initializers該類中的非靜態(tài)數(shù)據(jù)成員(包括嵌套類).

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 不在該列表中.

尾隨返回類型是聲明符的一部分 (§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 的可變參數(shù)版本不在其自己的trailing-return-type 范圍內(nèi)，并且無法通過名稱查找找到.

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，只需使用實(shí)際返回類型推導(dǎo)(并省略尾隨返回類型).在 C++11 中，你可以使用一個(gè)類模板和一個(gè)簡(jiǎn)單轉(zhuǎn)發(fā)的函數(shù)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...);
}

這篇關(guān)于gcc 可以編譯可變參數(shù)模板，而 clang 不能的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！