c++ - 如何在函数中返回模板容器(即 vector 、列表、数组)？

Question

我有以下类(class):

class Conversion {
//...
public:
    template<class T,
        template<class, class = std::allocator<T>> class> T doTheWork()
    {
        //do the work
        return {};
    }
};

我想将内容复制到顺序容器(vector、list、deque)，声明为:

template<class T, class Allocator = std::allocator<T>>

我对模板声明感到困惑。考虑到我想像下面的例子一样接收接收者，我应该如何声明copyContentToContainer？

例子:

int main() {
    Conversion conv;
    std::vector<std::string> container1 = conv.doTheWork();
    std::list<int> container2 = conv.doTheWork();
    std::deque<double> container3 = conv.doTheWork();
}

Answer 1

如果要分别指定元素类型和容器类型，可以

template<class T, template<class, class = std::allocator<T>> class C> 
C<T> doTheWork()
{
    //do the work
    return {};
}

然后

std::vector<std::string> container1 = conv.doTheWork<std::string, std::vector>();
std::list<int> container2 = conv.doTheWork<int, std::list>();
std::deque<double> container3 = conv.doTheWork<double, std::deque>();

否则，你可以

template<class T>
T doTheWork()
{
    //do the work
    return {};
}

然后

std::vector<std::string> container1 = conv.doTheWork<std::vector<std::string>>();
std::list<int> container2 = conv.doTheWork<std::list<int>>();
std::deque<double> container3 = conv.doTheWork<std::deque<double>>();

顺便说一句:模板参数不能是deduced从返回类型。它们只能从函数参数中推导出来。所以你必须明确指定它们。

When possible, the compiler will deduce the missing template arguments from the function arguments.

如果你不想将它们写两次，你可以应用 auto (C++11 起)。

auto container1 = conv.doTheWork<std::vector<std::string>>();
auto container2 = conv.doTheWork<std::list<int>>();
auto container3 = conv.doTheWork<std::deque<double>>();