成员初始化列表错误中的模板基构造函数调用
我有一个如下所示的基类:
I have a base class that looks like the following:
template<typename T>
class Base
{
public:
Base(int someValue);
virtual T someFunc() =0;
};
template<typename T>
Base<T>::Base(int someValue)
{}
然后是:
#include "base.hpp"
class Foo
: public Base<Foo>
{
public:
Foo(int someValue);
virtual Foo someFunc();
};
Foo::Foo(int someValue)
: Base(someValue)
{}
我从 gcc 4.2.1 得到以下错误.
I get the following error from gcc 4.2.1.
错误:Foo"类没有任何名为Base"的字段
error: class ‘Foo’ does not have any field named ‘Base’
我应该提到这在我运行 gcc 4.6.2 的 Fedora 机器上编译得很好.在我的 os x Lion 机器上编译时出现此错误.
I should mention this compiles fine on my Fedora box which is running gcc 4.6.2. This error occurs when compiling on my os x Lion machine.
编辑
问题似乎是我在调用构造函数时没有在 Foo 类中指明模板的类型.以下修复了 os x 中的错误.
Problem seems to be that I am not indicating type of template in the Foo class when calling the constructor. The following fixes the error in os x.
: Base<Foo>(someValue, parent)
编辑
是的,这确实看起来像一个错误.我之前提到的修复了 os x 下的错误,并且代码在 Fedora 中通过该修复可以正常编译.去看看os x中gcc有没有更新.
Yes this does look like a bug. What I mentioned before fixes the error under os x and code compiles fine in fedora with that fix. Will go and see if there is an update to gcc in os x.
推荐答案
第一:
[C++11: 12.6.2/3]:
mem-initializer-list 可以使用任何 class-or- 初始化基类decltype 表示该基类类型.
[C++11: 12.6.2/3]:
A mem-initializer-list can initialize a base class using any class-or-decltype that denotes that base class type.
[ 示例:
struct A { A(); };
typedef A global_A;
struct B { };
struct C: public A, public B { C(); };
C::C(): global_A() { } // mem-initializer for base A
――结束示例 ]
而且 Base
应该是一个有效的 injected-class-name 在这里的基础(也就是说,你可以用它代替 Base
):
And Base
should be a valid injected-class-name for the base here (that is, you can use it in place of Base<T>
):
[C++11: 14.6.1/1]:
像普通(非模板)类一样,类模板有一个注入类名(第 9 条).injected-class-name 可以用作 template-name 或 type-name. 当它与模板参数列表,作为模板模板参数的模板参数,或者作为中的最终标识符详细类型说明符在友元类模板声明中,它指的是类模板本身.否则,它等价于 template-name 后跟 template-parameters 包含在 <>
中的类模板的template-parameters.
[C++11: 14.6.1/1]:
Like normal (non-template) classes, class templates have an injected-class-name (Clause 9). The injected-class-name can be used as a template-name or a type-name. When it is used with a template-argument-list, as a template-argument for a template template-parameter, or as the final identifier in the elaborated-type-specifier of a friend class template declaration, it refers to the class template itself. Otherwise, it is equivalent to the template-name followed by the template-parameters of the class template enclosed in<>
.
[C++11: 14.6.1/3]:
类模板或类模板特化的 injected-class-name 可以用作template-name 或 type-name 范围内的任何位置.[ 示例:
[C++11: 14.6.1/3]:
The injected-class-name of a class template or class template specialization can be used either as a template-name or a type-name wherever it is in scope. [ Example:
template <class T> struct Base {
Base* p;
};
template <class T> struct Derived: public Base<T> {
typename Derived::Base* p; // meaning Derived::Base<T>
};
template<class T, template<class> class U = T::template Base> struct Third { };
Third<Base<int> > t; // OK: default argument uses injected-class-name as a template
――结束示例 ]
我没有发现任何内容表明这不适用于 ctor-initializer,所以我认为这是一个编译器错误.
I haven't found anything to indicate that this doesn't apply in the ctor-initializer, so I'd say that this is a compiler bug.
我的精简测试用例 在 GCC 4.1.2 中失败 和 GCC 4.3.4 但在 GCC 4.5.1 (C++11 模式).它似乎由 GCC bug 189 解决;在 GCC 4.5 发行说明中:
My stripped-down testcase fails in GCC 4.1.2 and GCC 4.3.4 but succeeds in GCC 4.5.1 (C++11 mode). It seems to be resolved by GCC bug 189; in the GCC 4.5 release notes:
G++ 现在实施 DR 176.以前 G++ 不支持使用模板基类的注入类名称作为类型名称,以及名称的查找发现模板的声明在封闭范围.现在查找名称找到注入的类名称,它可以用作类型或模板,具体取决于名称后面是否跟有模板参数列表.作为一个此更改的结果,以前接受的某些代码可能是格式错误,因为
G++ now implements DR 176. Previously G++ did not support using the injected-class-name of a template base class as a type name, and lookup of the name found the declaration of the template in the enclosing scope. Now lookup of the name finds the injected-class-name, which can be used either as a type or as a template, depending on whether or not the name is followed by a template argument list. As a result of this change, some code that was previously accepted may be ill-formed because
- 注入的类名不可访问,因为它来自私有库,或者
- 注入的类名不能用作模板模板参数的参数.
在这两种情况中的任何一种情况下,都可以通过添加嵌套名称说明符显式命名模板.第一个可以使用 -fno-access-control 解决;第二个只被拒绝与-迂腐.
In either of these cases, the code can be fixed by adding a nested-name-specifier to explicitly name the template. The first can be worked around with -fno-access-control; the second is only rejected with -pedantic.
<小时>
我用 Qt 抽象出来的精简测试用例:
My stripped-down testcase with Qt abstracted out:
template <typename T>
struct Base { };
struct Derived : Base<Derived> { // I love the smell of CRTP in the morning
Derived();
};
Derived::Derived() : Base() {};
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