什么是C ++的Java通配符? [英] What's a C++ equivalent of Java wildcards?

问题描述

在Java中，如果您有类似的类:

In Java if you have a class like:

class Box<E>
{
    some code
}

您可以使用通配符执行以下操作:

you can do the following using a wildcard:

Box<?> someBox;
someBox = new Box<Integer>();
someBox = new Box<Double>();

有没有办法在C ++中做到这一点?

Is there a way to do this in C++?

用更好的话来说，我该如何在C ++中声明一个变量，该变量可以容纳 Box< Integer> 或 Box< Double> 或 Box< WhateverDataTypeHere>?

In better words, how can I declare a variable in C++ that can hold either Box<Integer> or Box<Double> or Box<WhateverDataTypeHere>?

推荐答案

template< typename T>Box类应该从非模板基继承(例如 BasicBox类类).

template <typename T> class Box should inherit from a non-template base (let's say class BasicBox).

然后指向 BasicBox 的指针可以指向派生模板的专业化对象:

Then a pointer to BasicBox can point to objects of the specializations of the derived template:

BasicBox *someBox = new Box<int>;

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或者，因为在现代C ++™中应该避免手动管理内存，所以使用智能指针将是一个更好的主意:

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Or, since in modern C++™ manually managing memory should be avoided, using a smart pointer would be a better idea:

std::unique_ptr<BasicBox> someBox = std::make_unique<Box<int>>();

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