在std :: make_shared中使用c ++聚合初始化 [英] using c++ aggregate initialization in std::make_shared

问题描述

据我所知，以下代码构造了一个 Foo 类型的对象，然后将该对象移动到 std :: make_shared

Per my understanding, the following code constructs an object of type Foo and then moves that object into the memory allocated by std::make_shared

struct Foo
{
    std::string s;
    int i;
    char c;
};

int main(int argc, char* argv[])
{
    auto foo = std::make_shared<Foo>(Foo{"hello", 5, 'c' });
}

是否可以将初始化初始化 Foo 直接放入 std :: make_shared分配的内存 code>？

Is it possible to aggregate initialize Foo directly into the memory allocated by std::make_shared?

推荐答案

您可以创建一个带有可变构造函数模板的适配器来转发参数，例如：

You could create an adapter with a variadic constructor template to forward the arguments, something like:

template<class T>
struct aggregate_adapter : public T {
    template<class... Args>
    aggregate_adapter(Args&&... args) : T{ std::forward<Args>(args)... } {}
};

然后您可以执行以下操作：

And then you can do:

auto foo = std::make_shared<aggregate_adapter<Foo>>("hello", 5, 'c');

由于 aggregate_adapter< Foo> 和 Foo 是相关的， foo 可转换为 std :: shared_ptr< Foo>

Since aggregate_adapter<Foo> and Foo are related, foo is convertible to std::shared_ptr<Foo> as well.

不幸的是，使用转发还使不可能像 std :: make_shared< aggregate_adapter< Foo>>>>（{{h'，'e'，'l'，'l' ，'o'}，5，'c'）; 而不显式指定类型，但是对make_shared已经应用了相同的限制。

Unfortunately, the use of forwarding also makes it impossible to brace-init any of the members like std::make_shared<aggregate_adapter<Foo>>({'h','e','l','l','o'}, 5, 'c'); without specifying the type explicitly, but the same restriction applies to make_shared already.

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