如何避免使用特质类违反ODR [英] How to avoid violating ODR with traits classes

问题描述

从生产库在线阅读代码时，我发现了这样的内容

On reading code online from production libraries I found something like this

Traits.hpp

template <typename Type>
class Traits {
    template <typename T, 
              detail::EnableIfIsInstantiation<T, Type>* = nullptr>
    static void foo(T& object) { 
        object.foo();
    } 
};

SpecialTraits.hpp

template <>
class Traits<Special> {
    static void foo(Special& object) {
        object.foo();
    }
    static void foo(Special&& object) {
        object.special_foo();
    }
};

如果库实例化在一个翻译单元中将Traits用作Something的类型而不包含SpecialTraits.hpp的类型，然后实例化在另一翻译单元中使用特殊特性的类型，则将导致ODR违规.当这两个翻译单元链接在一起时，这会导致违反ODR.

This will cause an ODR violation if a library instantiates a type that uses Traits for Something in one translation unit without including SpecialTraits.hpp and then instantiates a type that uses the specialized traits in another translation unit. This would cause an ODR violation when those two translation units are linked together.

为避免此问题，建议的方法是什么?我是否必须求助于在原始Traits.hpp文件中包含所有专长?还有，如果不允许我编辑具有Special定义的文件怎么办?

What is the suggested way to avoid this problem? Do I have to resort to including all the specializations in the original Traits.hpp file? And what if I am not allowed to edit the file with the definition for Special?

注意，请忽略以下事实:在&&情况下，foo()可能由Special本身专门化了.我想不到一个更好的例子.

Note Please ignore the fact that foo() could have been specialized by Special itself in the && case. I could not think of a better example..

推荐答案

将专业名称放在"WidgetWrapper.hpp"中，而不是"Widget.hpp"中，并在各处都包含"WidgetWrapper.hpp".否则，请使用Boost提交错误报告，并期望它不会执行任何操作，因为确切的问题在15年前没有得到解决.

Put the specialization in "WidgetWrapper.hpp" instead of "Widget.hpp" and include "WidgetWrapper.hpp" everywhere. Otherwise, file a bug report with Boost and expect it to go nowhere since this exact problem was discussed 15 years ago with no resolution.

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