What is template metaprogramming? Why do I care about template metaprogramming? What is a typelist? These are the sorts of questions I asked myself when I first started looking at template metaprogramming seriously.

In this series of posts I will do my best to answer these questions using practical (from my line of work, anyway) examples that will hopefully convince you of the merits of template metaprograming (TMP). I will try to answer the first two questions together. TMP is writing code that is executed by the compiler and manipulates types rather than data. TMP is even more powerful when combined with constant expressions to do compile-time computations. I will restrict myself to using C++11 since this is currently best supported on the HPC systems used in academic settings. Currently the Intel compiler is what is prohibiting us from fully adopting C++14.

I will begin by discussing several runtime ideas that we need to implement at compile time in order to be able to have flexible metacode. TMP is functional programming in the sense that all metavariables or types and type aliases are immutable (const) and that all metafunctions return by value. In order to have flexible and useful metaprograms we want the idea of if, for and while, so let’s start with if. I will cover loops in part 2. How do we implement an if in TMP? Well we need a bool that can be true or false, so that could be one parameter to the metafunction. We also need the value that is returned if the bool is true and the return value if the bool is false. Okay, so our metafunction will need to take three parameters, so let’s start with that.

1
2
3
4
5
6
7
8
template <bool B, typename T, typename F>
struct if_c {
  using type = F;
};
template <typename T, typename F>
struct if_c<true, T, F> {
  using type = T;
};

What is happening here? We use a struct as the metafunction that applies the if and call it if_c (recall that if is a keyword). Arguments are passed to the metafunction as template parameters, and returned by setting a member type alias or static constexpr variable. For if_c the general (unspecialized) definition takes three template parameters: a bool, the metavalue (type) to be returned if the bool is true and the type to be returned if the bool is false. The general definition sets its member metavariable type to the false evaluation and we provide a (partial) template specialization on line 5 where the bool is true. The specialization sets the metavariable type to the true evaluation template parameter. How does this actually perform a conditional check? Well the compiler must use a specialization of if_c if it exists, if not it attempts to instantiate and use the general definition. Thus, if the first template parameter is true then the specialization is used, otherwise the general definition is used. Why did we choose to make the specialization for the true case? This is an arbitrary choice (and is the same one made in libc++). What we have implemented here is actually already in the standard library (stdlib, either libc++ or stdlibc++) and is named std::conditional.

Okay, great, we’re done with a compile time if! Wait, not so fast! Sometimes in runtime code we use if statements to avoid evaluating one branch altogether but that isn’t what happens with conditional. When using conditional all template parameters must be evaluable, which isn’t always what we need. Furthermore, maybe we want to “branch” compilation depending on a conditional or simply avoid the compiler checking a branch altogether so our compile time stays low. By branch I mean maybe we want a function that has the same name but operates on either associative containers (std::map or std::unordered_map) or sequence containers (std::array or std::vector). Well how do we do this? Clearly conditional isn’t the answer here, what we want is something that enables a (meta)function if a condition is met. This is exactly what std::enable_if does.

Let’s now implement this second type of if statement. Here is our implementation:

1
2
3
4
5
6
template <bool B, typename T = void>
struct enable_if {};
template <typename T>
struct enable_if<true, T> {
  using type = T;
};

While shorter than conditional it is more subtle in its utility. First enable_if takes two template parameters, a bool and a type where the type is void by default and really serves as the method of enabling a (meta)function. Note that the template parameter T has a default value of void for all specializations too. Just like arguments to functions may only have their default value specified once, template parameters may only have their default value specified once. We again have the general definition handle the false case and supply a specialization for the true case with a member metavariable type set to the template parameter T (void by default). So how do we actually use enable_if and how does it work? Well let’s first look at a practical use and then I’ll explain how enable_if works.

As our example let’s write a few type traits to check if a type is a map or unordered_map. The concepts are the same as what we used for conditional and enable_if and plays a central role in TMP. We first define a generic class template that inherits from std::false_type and therefore has a member variable named value whose value is (hopefully not surprisingly) false. Next we define a template specialization that is specialized for a map and inherits from std::true_type. This works because the compiler has to always choose the specialization if it can, just as with conditional and enable_if where the compiler selects the true specialization. Here are the implementations:

1
2
3
4
5
6
7
8
9
10
11
12
template <typename T>
struct is_std_map : std::false_type {};

template <typename K, typename V, typename Comp, typename A>
struct is_std_map<std::map<K, V, Comp, A>> : std::true_type {};

template <typename T>
struct is_std_unordered_map : std::false_type {};

template <typename K, typename V, typename Hash, typename KeyEqual, typename A>
struct is_std_unordered_map<std::unordered_map<K, V, Hash, KeyEqual, A>>
    : std::true_type {};

You may find it surprising that the specialization has more template parameters than the general definition. This is fine because the <std::map<K, V, Comp, A>> after the struct name on line 5 indicates that we have defined a specialization. Thus the compiler knows to first use this specialized definition if it is satisfied. Finally, what we have is that is_std_map<std::map<int, double>>::value is true while is_std_map<std::unordered_map<int, double>>::value is false. The implementation for unordered_map is essentially the same.

Since both map and unordered_map have the same dereferenced iterator (a pair) we might only care if the type is either a map or unordered_map but not which one. Let’s write another type trait that checks if a type is either map or unordered_map and call it is_map. Again the general definition will need the type to check as a parameter, and inherit from false_type. Now how do we use our above type traits to specialize is_map? We use enable_if, that’s how! In order to use enable_if we must add a second template parameter that we will not use other than to perform the enable_if check. Therefore we can use an unnamed template parameter with a default value, which we’ll set to void. The reason for setting a default value is so that the user of our type trait does not need to specify a second “filler” template parameter. Let me show you the general definition and specialization together, then explain the specialization.

1
2
3
4
5
6
template <typename T, typename = void>
struct is_map : std::false_type {};
template <typename T>
struct is_map<T, typename std::enable_if<is_std_map<T>::value or
                                         is_std_unordered_map<T>::value>::type>
    : std::true_type {};

This might seem like a bit to swallow, but let’s break it down. The specialization only takes one template parameter, the type T that we want to check. The second parameter holds the enable_if statement, which in this case takes the result of a logical or between the result of is_std_map<T>::value and is_std_unordered_map<T>::value. If T is either a map or an unordered_map then the specialization of enable_if is used and it has a member metavariable type. In this case the compiler can compile the code and must prefer the specialization over the general definition. If the result of the logical or is false then the general definition of enable_if is used and the compiler cannot compile the code because the general enable_if does not have a metavariable named type. As a result the compiler ignores the specialization of is_map and falls back to the generic definition. Lastly, the typename keyword is required to tell the compiler that type is a metavariable (a type) rather than a variable with a value.

So what have we discovered? Well we discovered that there are at least three different ways to perform if-else or switch statements at compile time. One is using std::conditional, another is using std::enable_if, and lastly we can also use (partial) template specialization to have the compiler select a certain definition over all others. Comparing these to runtime concepts conditional is probably most like the ternary operator, enable_if like if-else statements, and (partial) template specialization like if-else and switch statements.