可以在C ++ 03中有条件地声明friend类吗? [英] Can friend class be declared conditionally in C++03?

问题描述

仅当某些(编译时)条件为true时，我才想声明一个朋友类.例如:

I want to declare a friend class only if some (compile-time) condition is true. For example:

// pseudo-C++
class Foo {
    if(some_compile_time_condition) {
        friend class Bar;
    }
};

我在互联网上找不到任何解决方案.我仔细研究了在编译时动态生成结构的所有答案.他们中的许多人都使用C ++ 11 std::conditional，但是我想知道是否有可能在不使用预处理器的情况下在C ++ 03 中做到这一点.

I did not find any solution on the internet. I went through all the answers to the question Generating Structures dynamically at compile time. Many of them use the C++11 std::conditional, but I would like to know if it is possible to do this in C++03 without using the preprocessor.

此解决方案 https://stackoverflow.com/a/11376710/252576 将不起作用，因为friend ship不被继承(具有继承关系的朋友类).

This solution https://stackoverflow.com/a/11376710/252576 will not work because friendship is not inherited ( friend class with inheritance ).

编辑，只是为了使它更容易显示，如下面的注释中所述:此要求是不寻常的.这是我正在从事的硬件仿真新研究项目的一部分.测试平台是用C ++编写的，我想以波形形式显示变量.我研究了各种其他选择，并出于实际考虑，发现需要使用friend class.朋友将捕获这些值并生成波形，但是我更希望仅在需要波形时才与朋友互动，而不是一直与朋友互动.

Edit Just to make this more easily visible, as mentioned below in the comment: This requirement is unusual. This is part of a new research project in hardware simulation, that I am working on. The testbench is written in C++, and I want to display the variables in a waveform. I have researched various other options, and figured out that I need to use a friend class, due to practical considerations. The friend will capture the values and generate the waveform, but I would prefer to have the friend only when the waveform is required, and not all the time.

推荐答案

使用friend std::conditional<C, friendclass, void>::type;，其中C是您的条件.非类类型的朋友将被忽略.

Use friend std::conditional<C, friendclass, void>::type; where C is your condition. A nonclass type friend will be ignored.

条件模板很容易在C ++ 03中实现.但是，由于C ++ 03不支持typedef朋友，因此您需要在其中使用以下语法

The conditional template is easily implemented in C++03. However since C++03 does not support typedef friends you need to use the following syntax there

namespace detail { class friendclass {}; }

class Foo {
  friend class std::conditional<C, 
    friendclass, detail::friendclass>::type::friendclass;
};

请注意，在这种解决方法中，详细的虚拟类名必须与潜在朋友的名字匹配.

Note that the detail dummy class name needs to match the name of the potential friend in this workaround.

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