Compiler does not deduce template parameters (map std::vector -> std::vector)

Question

I have the following template.

template
std::vector map(const std::vector &v, std::function

        

Answer 1

Your function expects an std::function argument, but you're calling it with a lambda expression instead. The two are not the same type. A lambda is convertible to std::function, but template argument deduction requires exact matches and user defined conversions are not considered. Hence the deduction failure.

Deduction does work if you actually pass an std::function to map().

std::function<string(int const&)> fn = [] (int x) { return string(x,'X'); };
vector<string> strings = map(numbers, fn);

Live demo

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One possible workaround to avoid having to specify the template arguments is to modify the function to accept any kind of callable, rather than an std::function object.

template<typename T, typename Func>
std::vector<typename std::result_of<Func(T)>::type>
    map(const std::vector<T> &v, Func f) {
        // ...
    }

Another version of the same idea, using decltype and declval instead of result_of

template<typename T, typename Func>
std::vector<decltype(std::declval<Func>()(std::declval<T>()))>
    map(const std::vector<T> &v, Func f) {
        // ...
    }

Finally, using a trailing return type

template<typename T, typename Func>
auto map(const std::vector<T> &v, Func f) 
  -> std::vector<decltype(f(v[0]))> {
        // ...
    }

Live demo