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c++templatesc++11lambdastd-function
Yonghui
Yonghui

Reputation: 222

C++ 11 async programming using template

I'm writing a class to achieve task feature. Here is my code:

template<typename TType>
class YHMTask
{
public:
    YHMTask() {};
    std::future<TType> mTask;
};

template<typename T>
YHMTask<T> YHMCreateTask(std::function<T()> func)
{
    YHMTask<T> yhmTask;
    yhmTask.mTask = std::async(func);
    return yhmTask;
}

If I didn't use template in YHMCreateTask, I can use it like this:

YHMCreateTask([]{return 12;});

But when I writing this function using template. Compiler report this error:

error C2672: 'YHMCreateTask': no matching overloaded function found
error C2784: 'YHMTask<TType> YHMCreateTask(std::function<_Type(void)>)': could not deduce template argument for 'std::function<_Type(void)>' from 'main::<lambda_a66b482c6cd6dab7208879904592bde5>'

I have to use YHMCreateTask like this:

int TestFunc()
{
    return 11;
}
...
std::function<int(void)> funInt = TestFunc;
auto taskNew = YHMCreateTask(funInt);

I want YHMCreateTask can be used like YHMCreateTask([]{return 12;}); How should I do?

Upvotes: 1

Views: 317

Answers (1)

Jarod42
Jarod42

Reputation: 218228

You have to use something like:

template<typename F>
auto YHMCreateTask(F func)
-> YHMTask<decltype(func())>
{
    YHMTask<decltype(func())> yhmTask;
    yhmTask.mTask = std::async(func);
    return yhmTask;
}

T cannot be deduced from lamdba to construct std::function

Upvotes: 2

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