No matching function error when passing lambda function as argument

Question

I have a list of numbers.

I am trying to filter the list and only keep the positive numbers.

I am trying to do it by passing a lambda as an argument.

Answer 1

Change the function keep to

template<typename T, typename Func>
std::vector<T> keep(const std::vector<T> &original,
                    Func useful)
{
    // code as usual
}

Live example.

This works with an argument to useful being any one of these:

• lambda
• std::function
• functor
• function pointer

From the documentation:

The lambda expression constructs an unnamed prvalue temporary object of unique unnamed non-union non-aggregate type, known as closure type.

This means that two lambdas with the same code, would generate two different typed objects.

auto f1 = [](int) { return true; };
auto f2 = [](int) { return false; };
f2 = f1;                               // error: no viable '='

However, both of these are implicitly convert-able to the corresponding std::function types:

std::function<bool(int)> fn = f1;
fn = f2;

But then why doesn't it work in your case? This is because of template type deduction. Changing keep to

template<typename T>
std::vector<T> keep(const std::vector<T> &original,
                    std::function<bool(const int &)> useful)
// no type deduction for std::function's template, explicitly mentioned

will make your example compile without any cast at the caller site.

However, trying to match it against std::function<T> won't work since template type deduction doesn't consider any conversion. Template argument deduction looks for exact type matches. Implicit conversions don't matter at this stage. You've to explicitly cast it to a matching std::function as Atomic_alarm comments. Like Joseph says in How to convert a lambda to an std::function using templates:

Template type deduction tries to match the type of your lambda function to the std::function<T> which it just can't do in this case - these types are not the same. Template type deduction doesn't consider conversions between types.

While in the alternative solution what happens is something like this:

auto f = [](int i) { return (i >= 0); }

The type of f here is not std::function but some unnamed type deduced like it would for the template parameter Func above.

If you still want to do it the std::function way, see this answer which does it with an additional template indirection. See this answer and this post for related details.