In C++, std::optional<T> is a great way to represent a type that could hold a value of type T or nothing. However, it is somewhat clumsy to work with optional types when you want to chain operations on them, because you have to account for the nullopt case.

Inspired by the C# Elvis operator (also null-conditional operator) I set out to implement a pipe operator that allows us to chain optional expressions in a more expressive fashion. The case of an optional being nullopt is handled implicitly.

The Null-Conditional Operator in C#

There are two null-conditional operators in C# but I am only concerned by the null-conditional member access operator ?. which is also referred to as the Elvis operator. From the Microsoft Docs:

Available in C# 6 and later, a null-conditional operator applies a member access, ?., or element access, ?[], operation to its operand only if that operand evaluates to non-null; otherwise, it returns null.

This allows for a rather neat syntax where we can write code like this, e.g. for two String objects

String str2 = str1?.Insert(0, "preamble:")?.ToUpper();

In that case str2 will contain the upper case version of str1 with “PREAMBLE:” prepended, or null if str1 is null. The ability to chain expressions is where the real power of the null-condional operator comes into play: it allows for more expressive code without the need to check for null before each function call.

Using template metaprogramming techniques we can implement our own operator in C++ that does very similar things, conceptually. I want the oprator to apply functions to optionals. The class std::optional represents nullable types in C++ in a way that is in line with the ideas of functional programming.

Defining the Intended Behaviour in C++

I will use % as the operator in C++. It takes an opt of type std::optional<T> and applies a function to it. The C++ of the example above might look like this:

//std::optional<std::string> str1 = ... (given elsewhere)
std::optional<std::string> str2 = str1 % insert_at_begin("preamble:") % to_upper;

Where insert_at_begin and to_upper are functions that are defined appropriately. Lets have a look at the different kinds of functions that the operator can encounter.

Since I want to have my C++ implementation be a tool for functional programming, it is natural to talk about applying functions to optionals in a math-y way. As a notation I use \(f\) for a type of function, and T, U for C++ types, which are not void. I can think of the following types of functions we would want to apply to an object opt of type std::optional<T>:

1. \(f_1:\) T \(\rightarrow\) U.
2. \(f_2:\) T \(\rightarrow\) void
3. \(f_3:\) std::optional<T> \(\rightarrow\) std::optional<U>
4. \(f_4:\) std::optional<T> \(\rightarrow\) void
5. \(f_5:\) std::optional<T> \(\rightarrow\) U, where U is not an optional type.

There are two main types of functions: functions taking a T and those taking an std::optional<T>. We will discuss the intended behaviour separately.

Functions Taking T

The functions \(f_1\) and \(f_2\) know nothing of optionals. They might be implemented to work on type T, (const) T&, or T&&, but they have no way of dealing with optional input arguments. So this is what we have to implement.

We denote the operator by \(op\), which is itself a function which takes two parameters. Those two parameters are an optional and one of the function types defined above. The first case the easiest but it is instructive to think about it briefly: What do we want to happen when we apply function \(f_1:\) T \(\rightarrow\) U to an std::optional<T> using the C++ null-conditional operator? Well, in C# the operator just returns the result of the function or null if the object was null to begin with. We can do an analogous thing in C++: we always return an std::optional<U> instead of a U. The return value is nullopt if the input value was nullopt and otherwise is such that the wrapped value is \(u = f(\)t\()\) where t is the input argument of type T. The pseudocode looks like this:

std::optional<U> operator% (opt: optional<T> type, f: function type 1)
{
  if opt has value
  {
    return f(value of opt)
  }
  else
  {
    return nullopt
  }
}

The function signature of the % operator in this case is thus: \(op: (\)std::optional<T>\(,f_1)\rightarrow\)std::optional<U>. But what if the return type U is itself an optional type, i.e. U=std::optional<V>? In this case I do not want to wrap the return type of the function in a second optional. I will just have the operator return a type U instead.

So far so good. We can do a simple implementation for functions of type \(f_2\): Apply the (effectful) function only if the optional has a value. However, what do we return? We could return void, but I would like to be able to chain the operator and returning void would break this chain. So I would like to implement the operator like this (pseudocode):

std::optional<T>& operator% (opt: optional<T> type, f: function type 2)
{
  if opt thas value
  {
    apply f to value opt
  }
  return opt
}

The function signature of the optional operator in this case is \(op: (\)std::optional<T>\(,f_2)\rightarrow\)std::optional<T>. Functions of type \(f_2\) are purely effectful functions and should not modify its argument, because that would get us into all kinds of trouble in the implementation. More on that later. A possible function of type \(f_2\) would be to print to the console.

Functions Taking Optionals

For functions taking optionals, the behaviour of the operator should be different, because whoever designed the functions intended them to work on optionals. For cases 3,4 we can pretty much take the corresponding implementations from above. The difference is that the functions are always applied to the operand even if it is nullopt. The respective signatures are: \(op: (\)std::optional<T>\(,f_3)\rightarrow\)std::optional<U> and \(op: (\)std::optional<T>\(,f_4)\rightarrow\)std::optional<T>.

How about case number 5? Here the implementer expects an optional and guarantees a return type that is not an optional. There are also several ways to handle this. In the void return case I opted for a solution that was chainable and returns an optional. I am not going to do the same here. This function was designed to transform an optional to a different (non-optional) type. It is the end of a chain of calls. This is why I want to look the signature of the operator like this in this particular case: \(op: (\)std::optional<T>\(,f_5)\rightarrow\)U.

Summary

This is a quick summary of the function signatures of the C++ null-conditional operator % we developed above:

1. Depending on return type of \(f\) EITHER \(op(\)std::optional<T>\(,f_1)\rightarrow\)std::optional<U> if U is not an optional type, OR \(op(\)std::optional<T>\(,f_1)\rightarrow\)std::optional<V> if U is an optional type of V.
2. \(op(\)std::optional<T>\(,f_2)\rightarrow\)std::optional<T>
3. \(op: (\)std::optional<T>\(,f_3)\rightarrow\)std::optional<U>
4. \(op: (\)std::optional<T>\(,f_4)\rightarrow\)std::optional<T>
5. \(op: (\)std::optional<T>\(,f_5)\rightarrow\)U

To Be Continued

I will write a post on the implementation in the next part of the series. The function body of this operator is trivial but the tricky part of the implementation is getting the metaprogramming right.