Constructing std::function argument from lambda

问题

I have the following templated function (C++ latest standard is enabled in the compiler - but maybe 17 would be enough).

#include <functional>

template<typename TReturn, typename ...TArgs>
void MyFunction(const std::function<TReturn(TArgs...)>& callback);

int main()
{
    MyFunction(std::function([](int){}));
    MyFunction([](int){});
}

The first call compiles, when I explicitly convert it to std::function, but the second case does not.

In the first case the template deduction is done automatically, the compiler only knows that it shall convert it to some std::function and able to deduce the parameter and return type.

However in the second case it shall(?) also know that the lambda shall be converted to some std::function, but still unable to do it.

Is there a solution to get the second one running? Or can it be that for templates the automatic conversion does not take place at all?

The error message is:

error C2672: 'MyFunction': no matching overloaded function found

error C2784: 'void MyFunction(const std::function<_Ret(_Types...)> &)': could not deduce template argument for 'const std::function<_Ret(_Types...)>

note: see declaration of 'MyFunction'

What I am aiming for is a "python style decorator". So basically this:

template<typename TReturn, typename ...TArgs>
auto MyFunction(std::function<TReturn(TArgs...)>&& callback) -> std::function<TReturn(TArgs...)>
{
     return [callback = std::move(callback)](TArgs... args)->TReturn
     {
          return callback(std::forward<TArgs>(args)...);
    };
}

If I used a template instead of std::function, the how would I deduce the parameter pack and return value? Is there some way to get it from a callable via some "callable traits"?

回答1:

Or can it be that for templates the automatic conversion does not take place at all?

Yes. Implicit conversions won't be considered in template argument deduction.

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

That means given MyFunction([](int){});, the implicit conversion (from lambda to std::function) won't be considered, then the deduction for TReturn and TArgs fails and the invocation attempt fails too.

As the workarounds, you can

Use explicit conversion as you showed

As the comment suggested, just use a single template parameter for functors. e.g.

template<typename F>
auto MyFunction2(F&& callback)
{
     return [callback = std::move(callback)](auto&&... args)
     {
          return callback(std::forward<decltype(args)>(args)...);
     };
}

来源：https://stackoverflow.com/questions/54664303/constructing-stdfunction-argument-from-lambda

标签

c++

templates

lambda

stl

template-argument-deduction