I wrote a simple piping operator in c++. I just wanted to make sure that my code was robust, modern c++ code, and made correct use of perfect forwarding.

Here's the code:

#include <concepts>
#include <type_traits>
#include <utility>
#include <vector>
#include <iostream>
#include <deque>

//Accept an object on the left hand side and a function on the right
template <typename Any, typename... Args>
inline auto operator| (const Any& obj, std::invocable<Any> auto func){
    // Correct use of perfect forwarding??
    return func(std::forward<const Any&>(obj));

//Version for a functor which accepts a reference. Do I need to cover other cases?
template <typename Any, typename... Args>
inline auto operator| (Any& obj, std::invocable<Any&> auto func){
    return func(std::forward<Any&>(obj));

And here is how the user uses it:

int main() {
    auto x = std::vector{1 , 2 , 3 , 4 , 5 , 6};
    // Piping example 0
    auto y = 5 | [] (int x) {return x * 2;} 
               | [] (int x) {return x * 3;}
    std::cout << y << '\n';
    // Piping example 1
    x | [] (std::vector<int>& vec) -> std::vector<int>& {vec.push_back(7); return vec;};

    std::cout << x[6];
    return 0;

And if you want to clean up the lambda syntax (I don't mind personally) and have a cleaner, neater syntax you can do:

namespace piping {
    template <typename T>
    auto push_back(const T& obj) {
        return [obj] (auto& vec) -> auto& {vec.push_back(obj); return vec;};;

/* ... Some code */

int main() {
    auto x = std::vector{1 , 2 , 3 ,4 ,5 , 6};
    // A bit cleaner...
    x | piping::push_back(7);

Did I make correct use of modern c++ and perfect forwarding? Thanks in advance.


That seems a worthwhile objective. The syntax looks a bit clunky when we have to write lambdas inline like that, but you'll probably develop a library of useful filters (like standard Ranges ones) that look much more natural.

I don't think we've got the use of forwarding references quite right - and you'll be pleased to hear that fixing that should simplify the code a bit.

What we should do is use typename Any&&. Although that looks like an rvalue-reference, it's actually a forwarding reference because Any is a template parameter.

Any&& allows the template parameter to bind to either lvalue or rvalue, so we just need the one function:

template <typename Any>
auto operator| (Any&& obj, std::invocable<Any> auto&& func) {
    return func(std::forward<Any>(obj));

(Note also the auto&& for func, so it can be a mutable functor if we want).

Having written that, I'm not so convinced that I like operator|() to be modifying its argument. I would prefer a | b to leave a alone, and have to write a |= b when I intend a to be modified in place.

Minor points:

  • There are headers included but not needed for the template:
    #include <vector>
    #include <iostream>
    #include <deque>
  • inline is implicit for a template function.
  • We don't need to spell out the full return type for the "append 7" lambda in the example - -> auto& is easier.
  • \$\begingroup\$ Thank you for all your feedback and explaining perfect forwarding!! I had never quite understood it until now... About the "inline" keyword, the compiler (-O3) seemed to be better at optimising my code when the "inline" was there?? \$\endgroup\$ Feb 6 at 16:40
  • \$\begingroup\$ I'm surprised by that. If the inline is helpful, then continue using it. I don't think it can do any harm. \$\endgroup\$ Feb 6 at 16:41
  • \$\begingroup\$ It's worth searching the web for good articles on perfect forwarding and forwarding references. Another good search term is universal references, which is the older term for this construct. \$\endgroup\$ Feb 6 at 16:47
  • \$\begingroup\$ yes it surprised me as well. In fact both versions of my code inlined the function, but the version with "inline" straight up calculated everything at compile - time, while the version without inline had to call the lambda. \$\endgroup\$ Feb 6 at 16:48
  • \$\begingroup\$ OOh - perhaps it would make sense to add constexpr to the declaration? I'm not 100% sure about that, though - I'm just a beginner with constexpr. IIRC, the lambda can also be declared constexpr, but I'm really walking on thin ice now... \$\endgroup\$ Feb 6 at 16:54

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