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namespace { [[deprecated("This function has been replaced by combust()")]] void explode() = delete; // Using variaDic templates in reality to have all signatures covered void combust() {} } int main() { explode();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(); }
<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:11:2: error: call to deleted function 'explode': This function has been replaced by combust() explode(); ^~~~~~~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namespace { class [[deprecated("Class has been replaced by Bar")]] Foo { Foo () = delete; // And every other method I had in this class }; class Bar { }; } int main() { Foo f; Bar b; }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:13:5: warning: 'Foo' is deprecated: Class has been replaced by Bar [-Wdeprecated-declarations] Foo f; ^ 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:3:60: note: 'Foo' has been explicitly marked deprecated here class [[deprecated("Class has been replaced by Bar")]] Foo ^ 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class [[deprecated("Class has been replaced by Bar")]] Foo = delete;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struct [[deprecated("Class has been replaced by Bar")]] Foo { Foo() = delete; };
struct [[deprecated("Class has been replaced by Bar")]] Foo;
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 void explode() { static_assert(I && false,"This function has been replaced by combust()"); } int main() { // error: static_assert Failed: "This function has been replaced by combust()" explode(); }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 class FooDeprec { static_assert(I && false,"Class has been replaced by Bar"); FooDeprec() = default; // no need to delete }; using Foo = FooDeprec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; } int main() { // error: static_assert Failed "Class has been replaced by Bar" Foo f; }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.
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