概念std :: derived_from当参数是一个明智的指针时

发布于 2025-01-31 12:36:05 字数 363 浏览 3 评论 0原文

我有类似的功能

bool RegisterModel (std::shared_ptr<DerivedA> model) { }

bool RegisterModel (std::shared_ptr<DerivedB> model) { }

，我想利用C ++ 20概念并这样实现：

bool RegisterModel (std::derived_from<BaseClass> auto model) { }

这是不起作用的，因为我传递了共享的指针。有可能需要一个共享指针，该指针容纳baseclass的对象？

原文

I have a few functions like so

bool RegisterModel (std::shared_ptr<DerivedA> model) { }

bool RegisterModel (std::shared_ptr<DerivedB> model) { }

and i would like to make use of c++ 20 concepts and implement it like this:

bool RegisterModel (std::derived_from<BaseClass> auto model) { }

This does not work, because i'm passing in shared pointers. It is somehow possible to require a shared pointer that holds an object derived from BaseClass?

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冰魂雪魄 2025-02-07 12:36:05

从std :: shared_ptr＆lt; t＆gt;中推导t>

template<std::derived_from<BaseClass> T>
bool RegisterModel (std::shared_ptr<T> model) { }

// Or as an abbreviated function template
bool RegisterModel (std::shared_ptr<std::derived_from<BaseClass> auto> model) { }

Deduce the T from a std::shared_ptr<T> and constrain that:

template<std::derived_from<BaseClass> T>
bool RegisterModel (std::shared_ptr<T> model) { }

// Or as an abbreviated function template
bool RegisterModel (std::shared_ptr<std::derived_from<BaseClass> auto> model) { }

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