C++ Truths

In the last post we looked at basic usage of C++17 Fold Expressions. I found that many posts on this topic discuss simple types and ignore how folds may be applicable to more complex types as well. [Edit: Please see the comments section for some examples elsewhere in the blogosphere.] In this post I'm going to describe folding over functions. Composing Functions Function composition is a powerful way of creating complex functions from simple ones. Functions that accept a single argument and return a value are easily composable. Consider the following example to compose two std::functions. template <class A, class B, class C> std::function<C(A)> compose(std::function<B(A)> f, std::function<C(B)> g) { return [=](A a) -> C { return g(f(a)); }; } int main(void) { std::function<int(std::string)> to_num = [](std::string s) { return atoi(s.c_str()); }; std::function<bool(int)> is_even = [](int i) { return i%2==0; }; auto is_str_e...

C++17 has an interesting new feature called fold expressions . Fold expressions offer a compact syntax to apply a binary operation to the elements of a parameter pack. Here’s an example. template <typename... Args> auto addall(Args... args) { return (... + args); } addall(1,2,3,4,5); // returns 15. This particular example is a unary left fold . It's equivalent to ((((1+2)+3)+4)+5). It reduces/folds the parameter pack of integers into a single integer by applying the binary operator successively. It's unary because it does not explicitly specify an init (a.k.a. identity) argument. So, let add it. template <typename... Args> auto addall(Args... args) { return (0 + ... + args); } addall(1,2,3,4,5); // returns 15. This version of addall is a binary left fold . The init argument is 0 and it's redundant (in this case). That's because this fold expression is equivalent to (((((0+1)+2)+3)+4)+5). Explicit identity elements will come in handy a little ...