将 lambda 绑定到函数以获取返回结果

Question

尝试将 lambda 作为参数传递给采用 std::function 的函数，然后使用类型推导获取返回值。但是编译失败。

#include <future>


class WorkQue
{
    public:
        template<typename R, typename ...Args>
        std::future<R> addItem(std::function<R(Args...)> task, Args... args)
        {
            std::promise<R>     promise;
            std::future<R>      future = promise.get_future();

            // STUFF

            return future;
        }
};

int main()
{
    WorkQue     que;
    int         x = 1;
    int         y = 2;

    // This fails
    que.addItem([](int x, int y){return x+y;}, x, y);

    // I can see needing to specify the return type.
    // But even this does not work.
    // que.addItem<int>([](int x, int y){return x+y;}, x, y);

    // This works fine.
    // But I was hoping not to need to specify this every time.
    //que.addItem<int>(std::function<int(int,int)>([](int x, int y){return x+y;}), x, y);
}

编译错误是：

> g++ -std=c++1y thread.cpp
thread.cpp:24:9: error: no matching member function for call to 'addItem'
    que.addItem([](int x, int y){return x+y;}, x, y);
    ~~~~^~~~~~~
thread.cpp:9:24: note: candidate template ignored: could not match 'function<type-parameter-0-0 (type-parameter-0-1...)>' against '<lambda at thread.cpp:24:17>'
        std::future<R> addItem(std::function<R(Args...)> task, Args... args)
                       ^
1 error generated.

Answer 1

你不需要使用std::function

class WorkQue
{
    public:
        template<typename Func, typename ...Args>
        auto addItem(Func && task, Args &&... args) -> std::future<decltype(task(std::forward<Args>(args)...))>
        {
            using R = decltype(task(std::forward<Args>(args)...));
            std::promise<R>     promise;
            std::future<R>      future = promise.get_future();

            // STUFF

            return future;
        }
};

http://coliru.stacked-crooked.com/a/c1a8b02a1fbf5123