Pass lambda as template function parameter

Question

Why doesn't the following code compile (in C++11 mode)?

#include <vector>

template<typename From, typename To>
void qux(const std::vector<From>&, To (&)(const From&)) { }

struct T { };

void foo(const std::vector<T>& ts) {
    qux(ts, [](const T&) { return 42; });
}

The error message is:

prog.cc:9:5: error: no matching function for call to 'qux'
    qux(ts, [](const T&) { return 42; });
    ^~~
prog.cc:4:6: note: candidate template ignored: could not match 'To (const From &)' against '(lambda at prog.cc:9:13)'
void qux(const std::vector<From>&, To (&)(const From&)) { }
     ^

But it doesn't explain why it couldn't match the parameter.

If I make qux a non-template function, replacing From with T and To with int, it compiles.

Answer 1

A lambda function isn't a normal function. Each lambda has its own type that is not To (&)(const From&) in any case.
A non capturing lambda can decay to To (*)(const From&) in your case using:

qux(ts, +[](const T&) { return 42; });

---

As noted in the comments, the best you can do to get it out from a lambda is this:

#include <vector>

template<typename From, typename To>
void qux(const std::vector<From>&, To (&)(const From&)) { }

struct T { };

void foo(const std::vector<T>& ts) {
    qux(ts, *+[](const T&) { return 42; });
}

int main() {}

---

Note: I assumed that deducing return type and types of the arguments is mandatory for the real problem. Otherwise you can easily deduce the whole lambda as a generic callable object and use it directly, no need to decay anything.

Answer 2

If you don't need to use the deduced To type, you can just deduce the type of the whole parameter:

template<typename From, typename F>
void qux(const std::vector<From>&, const F&) { }