C unique_ptr从派生到基础铸造

C++ unique_ptr casting from derived to base

本文关键字:派生 unique ptr      更新时间:2023-10-16

我一直在阅读,并努力理解和实现此功能。

我有一个基础班级,是老师&学生继承。我想将它们都放在"人"类型的矢量中。我尝试了一些事情。但是,我不断遇到错误页面,我正在努力理解它们。

此当前代码用G -STD = C 17 Test.cpp编译在给我:

Undefined symbols for architecture x86_64:
  "Person::~Person()", referenced from:
      Teacher::~Teacher() in test-9423bf.o
      Student::~Student() in test-9423bf.o
ld: symbol(s) not found for architecture x86_64

会感谢简单地写的有关C 功能的任何提示和良好的参考。

#include <iostream>
#include <vector>
#include <memory>
class Person {
public:
    virtual void printName() = 0;
    virtual ~Person() = 0;
};
class Teacher : public Person {
public:
    void printName() {
        std::cout << "Hello My Name is Teacher" << std::endl;
    }
    ~Teacher() {}
};
class Student : public Person {
public:
    void printName() {
        std::cout << "Hello My Name Is Student" << std::endl;
    }
    ~Student() {}
};
//Capturing the raw pointer and letting it go out of scope
template<typename Person, typename Teacher>
std::unique_ptr<Person> static_unique_pointer_cast (std::unique_ptr<Teacher>&& old){
    return std::unique_ptr<Person>{static_cast<Person*>(old.release())};
    //conversion: unique_ptr<FROM>->FROM*->TO*->unique_ptr<TO>
}
auto main() -> int {

    auto t1 = std::make_unique<Teacher>();
    auto t2 = std::make_unique<Teacher>();
    auto t3 = std::make_unique<Teacher>();
    auto s1 = std::make_unique<Student>();
    auto s2 = std::make_unique<Student>();
    auto s3 = std::make_unique<Student>();
    std::vector<std::unique_ptr<Person>> v;
    // v.push_back(static_unique_pointer_cast<Person>(std::move(s1)));
    auto foo = static_unique_pointer_cast<Person>(std::move(s1));
    // std::vector<std::unique_ptr<Person>> ve = {
    //     std::move(t1), 
    //     std::move(t2), 
    //     std::move(t3), 
    //     std::move(s1), 
    //     std::move(s2), 
    //     std::move(s3)
    //     };
    return 0;
}

编辑:我通过将基类驱动器更改为默认值来使它起作用。

我现在有一个:

std::vector<std::unique_ptr<Person>> v;
v.push_back(static_unique_pointer_cast<Person>(std::move(s1)));
v.push_back(static_unique_pointer_cast<Person>(std::move(s1)));
for (auto item: v) {
    item->printName();
}

但是我收到以下错误:

error: call to implicitly-deleted copy constructor of 'std::__1::unique_ptr<Person, std::__1::default_delete<Person> >'
    for (auto item: v) {

编辑2:

我使用时上述有效:

for (auto &&item: v) {
        item->printName();
    }

有人可以向我解释一下吗?矢量具有独特的指针(曾经是rvalue(特别是Exrvalue),但现在不是。为什么我需要使用自动&amp;?

我已经清理了代码,请尝试解释更改。

#include <iostream>
#include <vector>
#include <memory>
class Person {
public:
    virtual ~Person() = default; // before this was a pure-virtual d'tor. Usually, you don't need it, but the linker told you it wasn't implemented
    virtual void printName() = 0;
};
class Teacher : public Person {
public:
    void printName() override {
        std::cout << "Hello My Name is Teachern";
    }
    // removed d'tor since you already have a default virtual destructor here
};
class Student : public Person {
public:
    void printName() override {
        std::cout << "Hello My Name Is Studentn";
    }
    // removed d'tor since you already have a default virtual destructor here
};
// removed template this upcasting is almost straight forward.
auto main() -> int {
    auto t1 = std::make_unique<Teacher>();
    auto t2 = std::make_unique<Teacher>();
    auto t3 = std::make_unique<Teacher>();
    auto s1 = std::make_unique<Student>();
    auto s2 = std::make_unique<Student>();
    auto s3 = std::make_unique<Student>();
    std::vector<std::unique_ptr<Person>> v;
    v.push_back(std::move(t1));
    v.push_back(std::move(t2));
    v.push_back(std::move(t3)); // Easy to add Students and Teachers
    v.push_back(std::move(s1));
    v.push_back(std::move(s2));
    v.push_back(std::move(s3));
    for (auto &item: v) { // Taking a reference, `&`, the pointers in `v` still stay there after the loop and can be reused. Don't use `&&` unless you want the use the pointers once only.
        item->printName();
    }
    return 0;
}