C ++ 11中的高阶函数

Question

I'm trying to write a generic fold function using the new anonymous functions available in C++11, here is what I have:

template<typename T>
T foldl(std::function<T(T,T)> f, T initial, std::vector<T> items) {
    T accum = initial;
    for(typename std::vector<T>::iterator it = items.begin(); it != items.end(); ++it) {
        accum = f(accum, (*it));
    }
    return accum;
}

The following attempt to use it:

std::vector<int> arr;
arr.assign(8, 2);
foldl([] (int x, int y) -> int { return x * y; }, 1, arr);

causes an error:

main.cpp:44:61: error: no matching function for call to 'foldl(main(int, char**)::<lambda(int, int)>, int, std::vector<int>&)'
main.cpp:44:61: note: candidate is:
main.cpp:20:3: note: template<class T> T foldl(std::function<T(T, T)>, T, std::vector<T>)
main.cpp:20:3: note:   template argument deduction/substitution failed:
main.cpp:44:61: note:   'main(int, char**)::<lambda(int, int)>' is not derived from 'std::function<T(T, T)>'

It seems to me that using std::function is not the right way to go about defining the type of f . How can I correct this?

Answer 1

Your code is not very generic. There is no need to require a function , vector , or anything of the kind. And generally, in C++, functions would go at the end of the argument list (especially important for lambdas, as they can be big).

So it would be better (ie: more standard) to write this as this:

template<typename Range, typename Accum>
typename Range::value_type foldl(const Range &items, const typename Range::value_type &initial, Accum f)
{
    typename Range::value_type accum = initial;
    for(const auto &val : items) {
        accum = f(accum, val);
    }

    return accum;
}

Or you could just use std::accumulate which does the exact same thing .

Answer 2

I'm not certain why this template is failing, but switching over to using a template parameter for the function instead of std::function<> seems to work wonderfully.

template<typename T, typename F>
T foldl(F f, T initial, std::vector<T> items) {

Answer 3

If you simply convert it to an std::function first, and then use that, it will work:

std::vector<int> arr;
arr.assign(8, 2);
std::function<int(int,int)> f = [] (int x, int y) -> int { return x * y; };
foldl(f, 1, arr);

The problem is that lambda is a different type than std::function . Each lambda is a separate type. Although a lambda (and all other function objects) is convertible to a std::function with appropriate type parameter, the compiler has no idea what template argument for T would make it convertible. (We happen to know that T = int would work, but there is no general way to figure it out.) So it cannot compile it.