std::function template argument resolution

Question

I am currently working on a library where I am chaining function objects.

I am creating a function template that takes a callable object (std::function at the moment) and is parametrized on the output and input type of the function. Here is a simplified version of what I am defining:

template <typename In, typename Out>
std::vector<Out> process(std::vector<In> vals, std::function< Out(In) > func)
{
    // apply func for each value in vals
    return result;
}

The problem I am having is on usage. It seems that when I pass a lambda, the compiler cannot deduce the type correctly, so complains that the function doesn't exist:

std::vector<string> strings;
// does NOT compile
auto chars = process(strings,
        []( std::string s ) -> char
        {
            return s[0]; // return first char
        }
);

If I explicitly wrap the lambda in std::function, the program compiles:

std::vector<string> strings;
// DOES compile
auto chars = process(strings,
        std::function< char(std::string) >(
        []( std::string s ) -> char
        {
            return s[0]; // return first char
        })
);

I haven't tested passing function pointers or function objects yet, but it seems like it will be difficult for the compiler to deduce the the In and Out parameters if I'm not directly passing the explicit std::function object.

My question is this: is there a way to get around this, so that I can deduce the input/return type of a callable object without explicitly mentioning them at the call site?

Perhaps parametrize the template on the function type instead of the input/return types? Essentially I need to deduce the In and Out type for an arbitrary callable. Perhaps some kind of auto/decltype trick for the return type of the template function?

Thank you.

Answer 1

I think what you can do is to create an intermediate return type deducing function which uses decltype to determine the arguments to be passed to the actual function object:

template <typename Out, typename In>
std::vector<Out> process_intern(std::vector<In> vals, std::function< Out(In) > func)
{
    // whatever
}

template <typename In, typename Func>
auto process(std::vector<In> vals, Func func) -> std::vector<decltype(func(vals[0]))>
{
    return process_intern<decltype(func(vals[0]))>(vals, func);
}

Of course, you may want to consider implementing the logic in process() directly anyway unless there is a reason to type erase the function type.

Answer 2

is there a way to get around this, so that I can deduce the input/return type of a callable object without explicitly mentioning them at the call site?

No. User-defined conversions are not considered in template argument deduction. The compiler have to come up with In and Out such that the type of the parameter and the type of the argument have to match (almost) exactly, but they never will in this case.

Perhaps parametrize the template on the function type instead of the input/return types

Yes, that's what is normally done (take a look at the standard library algorithms, for example)