c – 如何在保持弃用警告的同时删除类

发布时间：2020-12-16 07:20:21 所属栏目：百科来源：网络整理

导读：我正在尝试寻找一种从我的库中删除已弃用的类的好方法,同时保留好的错误消息.这个想法是基于我已经完成的功能： namespace{ [[deprecated("This function has been replaced by combust()")]] void explode() = delete; // Using variadic templates in real

我正在尝试寻找一种从我的库中删除已弃用的类的好方法,同时保留好的错误消息.这个想法是基于我已经完成的功能：

namespace
{
    [[deprecated("This function has been replaced by combust()")]]
    void explode() = delete; // Using variadic templates in reality to have all signatures covered

    void combust() {}
}

int main()
{
    explode();
    combust();
}

在clang中,这给了我一个很好的错误信息：

<source>:11:2: error: call to deleted function 'explode': This function has been replaced by combust()
explode();
^~~~~~~

GCC仅向我提供此功能已被删除的消息.虽然,这仍然表明每个人都试图升级忽略弃用警告的库的意图.

所以,因为这是一个c库,我主要有类,我正在寻找正确的方法来做类似这些类.我目前的方法如下：

namespace
{
    class [[deprecated("Class has been replaced by Bar")]] Foo
    {
        Foo () = delete; // And every other method I had in this class
    };
    class Bar
    {
    };
}

int main()
{
    Foo f;
    Bar b;
}

这基本上在clang中给出了以下警告/错误(在GCC中类似)：

<source>:13:5: warning: 'Foo' is deprecated: Class has been replaced by Bar [-Wdeprecated-declarations]
Foo f;
^
<source>:3:60: note: 'Foo' has been explicitly marked deprecated here
class [[deprecated("Class has been replaced by Bar")]] Foo
^
<source>:13:9: error: call to deleted constructor of '(anonymous namespace)::Foo'
Foo f;
^
<source>:5:8: note: 'Foo' has been explicitly marked deleted here
Foo () = delete;
^

我可以使用这个无用的函数代码,因为这是单行,这对于类来说变得很麻烦,因为它们可以有很多方法.

所以,我正在寻找的是一个很好的方法来做到以下几点:(非编译代码)

class [[deprecated("Class has been replaced by Bar")]] Foo = delete;

封闭的我得到了一个oneliner是：

struct [[deprecated("Class has been replaced by Bar")]] Foo { Foo() = delete; };
struct [[deprecated("Class has been replaced by Bar")]] Foo;

请注意,这不包括通过引用传递Foo并调用某些方法的情况.

有没有更好的解决方案来删除类,同时对以下某些版本有明确的弃用警告？

解决方法

可以使用static_assert执行此操作,因为它会导致带有消息的编译时错误.但是,如果置于非模板化函数中,它将始终置位.为了解决这个问题,我们将其作为模板函数,断言依赖于参数.如果不使用该函数,这不会导致错误,因为它不会被实例化.

template <int I = 0>
void explode()
{
    static_assert(I && false,"This function has been replaced by combust()");
}

int main()
{
    // error: static_assert failed: "This function has been replaced by combust()"
    explode();
}

这也可以用于课程.但是,由于需要模板参数,因此需要使用typedef.

namespace
{
    template <int I = 0>
    class FooDeprec
    {
        static_assert(I && false,"Class has been replaced by Bar");

        FooDeprec() = default; // no need to delete
    };

    using Foo = FooDeprec<>;
}

int main()
{
    // error: static_assert failed "Class has been replaced by Bar"
    Foo f;
}

这样做的好处是不需要进行太多更改 – 您可以保留成员函数的声明.

（编辑：李大同）

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