英文:
C++ type of item
问题
以下是您的代码的翻译部分：
using namespace std;class SimpleProduct {char look = 'x';string name = "Undefined";string type = "Undefined";string description = "Undefined";public:char getLook() const {return look;}string getType() const {return type;}string getName() const {return name;}string getDescription() const {return description;}SimpleProduct(char look = 'x', string &&name = "Undefined", string &&type = "Undefined", string &&description = "Undefined");virtual string toString() const;};class TallProduct : public SimpleProduct {public:TallProduct(char look, string &&name = "Undefined", string &&type = "Undefined", string &&description = "Undefined");string toString() const;};所有产品都保存在另一个类的属性 `unique_ptr<Product> ***products;` 中，我想检查某个特定产品是否是 Product 或 TallProduct 类型。例如：```cppif (y != 0 && products[x][y-1][position] is type TallProduct) { // 伪代码return 0;}
英文:
Here is my code:
using namespace std;class SimpleProduct {char look = 'x';string name = "Undefined";string type = "Undefined";string description = "Undefined";public:char getLook() const {return look;}string getType() const {return type;}string getName() const {return name;}string getDescription() const {return description;}SimpleProduct(char look = 'x', string &&name = "Undefined", string &&type = "Undefined", string &&description = "Undefined");virtual string toString() const;};class TallProduct : public SimpleProduct {public:TallProduct(char look, string &&name = "Undefined", string &&type = "Undefined", string &&description = "Undefined");string toString() const;};
All of products are saved in attribute unique_ptr<Product> ***products; of another class and I want to check, whether some specific product is type Product or TallProduct.
So for example:
if (y != 0 && products[x][y-1][position] is type TallProduct) { //pseudocodereturn 0;}
答案1
得分: 3
如果你使用 dynamic_cast<>，你可以检查类型是否是特定的，如果它是继承的并且是多态的。
这是一个小示例：
class A {public:virtual ~A() {};};class B : public A {};class C : public A {};int main() {A* b = new B();C* c = new C();std::cout << (dynamic_cast<B*>(b) != nullptr) << "\n";std::cout << (dynamic_cast<B*>(c) != nullptr) << "\n";}
这会产生以下输出：
10
这意味着 b 的类型是 B，但 c 的类型不是。
英文:
If you use dynamic_cast<>, you can check if type is something specific if it is inherited and it is polymorphic.
Here is a small sample:
class A {public:virtual ~A() {};};class B : public A {};class C : public A {};int main() {A* b = new B();C* c = new C();std::cout << (dynamic_cast<B*>(b) != nullptr) << "\n";std::cout << (dynamic_cast<B*>(c) != nullptr) << "\n";}
which creates following output:
10
That means b is from type B but c is not.
答案2
得分: 0
为使此工作正常，您需要执行两项操作：
- 在您的SimpleProduct中至少添加一个虚函数，以便层次结构有一个虚函数表（vtable）。您需要在这里添加虚拟析构函数，以使
unique_ptr能够正常工作：virtual ~SimpleProduct() = default - 使用dynamic_cast的指针变体来检查转换，即
auto tallProduct = dynamic_cast<TallProduct*>(products[x][y-1][position]);如果指针不是nullptr，则可以继续进行
请注意，这通常被认为是不良风格，因为它违反了面向对象编程的“告诉，不要询问”原则。
英文:
To get this to work, you need to do two things:
- Add at least one virtual function to your SimpleProduct, so that the hierarchy gets a vtable. You want the virtual destructor here, so that
unique_ptrwill do it's job properly:virtual ~SimpleProduct() = default - Check the conversion with the pointer-variant of dynamic_cast, i.e.
auto tallProduct = dynamic_cast<TallProduct*>(products[x][y-1][position]);You can proceed if the pointer is notnullptr
Note that this is generally considered bad style, because it violates the "Tell, don't ask" mantra of OOP
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