我正在尝试结合使用 C++ 11 直接数据成员初始化和“使用”语法来继承基类的构造函数。现在使用 gcc 5.4.0(在 Ubuntu 16.04 上)我观察到一个奇怪的错误，如果数据成员类型没有默认构造函数。查看以下最小示例可能最容易理解:

#include <iostream>

struct Foo {
  Foo(int arg) { std::cout << "Foo::Foo(" << arg << ")" << std::endl; }
};

struct Base {
  Base(int arg) { std::cout << "Base::Base(" << arg << ")" << std::endl; }
};

struct Derived : public Base {
  using Base::Base;
  Foo foo{42};
};

int main() {
  Derived derived{120};
}

此代码使用 clang 以预期的行为编译和执行。使用 gcc 它不会编译，因为编译器会删除构造函数 Derived::Derived(int):

ttt.cpp: In function ‘int main()’:
ttt.cpp:17:22: error: use of deleted function ‘Derived::Derived(int)’
   Derived derived{120};
                      ^
ttt.cpp:12:15: note: ‘Derived::Derived(int)’ is implicitly deleted because the default definition would be ill-formed:
   using Base::Base;
               ^
ttt.cpp:12:15: error: no matching function for call to ‘Foo::Foo()’
ttt.cpp:4:3: note: candidate: Foo::Foo(int)
   Foo(int arg) { std::cout << "Foo::Foo(" << arg << ")" << std::endl; }
   ^
ttt.cpp:4:3: note:   candidate expects 1 argument, 0 provided
ttt.cpp:3:8: note: candidate: constexpr Foo::Foo(const Foo&)
 struct Foo {
        ^
ttt.cpp:3:8: note:   candidate expects 1 argument, 0 provided
ttt.cpp:3:8: note: candidate: constexpr Foo::Foo(Foo&&)
ttt.cpp:3:8: note:   candidate expects 1 argument, 0 provided

如果我像这样向 Foo 添加一个默认构造函数:

  Foo() { std::cout << "Foo::Foo()" << std::endl; };

gcc 也可以编译它。代码的行为方式完全相同，特别是添加的 Foo 的默认构造函数永远不会被执行。

所以我现在的问题是，这是有效的 C++ 11 吗？如果是，我可能在 gcc 中发现了一个错误。否则，gcc 和 clang 不应该都给我一条错误消息，指出这不是有效的 C++ 11？

在@vlad-from-moscow 很好地回答了这个问题后编辑:这个错误似乎也存在于 gcc 6.2 中，所以我将提交一个错误报告。

第二次编辑:已经有一个错误，我在第一次搜索中没有找到:https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67054

最佳答案

gcc 不满足 C++ 标准。 Derived 类的继承构造函数应使用为 Derived 继承构造函数指定的参数调用其 mem-initializer 列表中的 Base 构造函数。

有写在C++标准(12.9继承构造函数)

8 An inheriting constructor for a class is implicitly defined when it is odr-used (3.2) to create an object of its class type (1.8). An implicitly-defined inheriting constructor performs the set of initializations of the class that would be performed by a user-written inline constructor for that class with a mem-initializer-list whose only mem-initializer has a mem-initializer-id that names the base class denoted in the nested-name-specifier of the using-declaration and an expression-list as specified below, and where the compound-statement in its function body is empty (12.6.2). If that user-written constructor would be ill-formed, the program is ill-formed. Each expression in the expression-list is of the form static_cast(p), where p is the name of the corresponding constructor parameter and T is the declared type of p.

也根据章节(12.6.2 初始化基和成员)

8 In a non-delegating constructor, if a given non-static data member or base class is not designated by a mem-initializer-id (including the case where there is no mem-initializer-list because the constructor has noctor-initializer) and the entity is not a virtual base class of an abstract class (10.4), then

— if the entity is a non-static data member that has a brace-or-equal-initializer, the entity is initialized as specified in 8.5;

关于c++ - 构造函数继承和直接成员初始化，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40932844/