C++20 introduces many new features, but I want to focus on one in particular in this post.

Non-Type Template Parameters

C++ has long supported non-type template parameters:

template <int I>
struct foo {};

We can then instantiate this class with any int:

foo<1729> f;

This is very unsurprising. One extremely important aspect of template specializations is that any two specializations that have equal template arguments refer to the same entity, and two specializations with any inequal template arguments refer to separate entities. That is: foo<1> and foo<2> are separate types, but foo<1> in one point in a program is the same type as any other mention of foo<1>, even if that template argument is not a literal 1, but any constant expression that evaluates to 1:

constexpr int square(int n) {
    return n * n;

foo<49>        a;
foo<square(7)> b;
foo<49>& ref = b; // Okay!

Here, foo<square(7)> is the same type as foo<49>, because the template arguments have the same value as constant expressions.

Classes Types as Template Parameters?

Suppose I have a simple class:

struct point {
    int x;
    int y;

Can I use a point as a template parameter?

template <point P>
struct location {};

Before C++20, the answer was a resounding no. The issue lies in having a compile-time definition of whether any two point objects are “equal.” It is the job of the compiler and linker together to enforce this one-definition rule, and until C++20 we were unable to come up with a good way to define equivalence in such a case. While we could provide a constexpr operator==, it is possible to create a constexpr bool operator==(point) { return true; }, which would tell the compiler that all points are equal, but this is confusing and nonsensical, and would require that we load a constexpr-evaluator at the link phase.

Instead, in C++20, we allow class-type non-type template parameters if such a type meets certain constraints. I won’t elaborate on those here, but we can use (among other things) any class-type whose members are either (1) valid as non-type template parameters, or (2) an array of such types. This very simple rule gives us enough lee-way to define fixed_string.

Meet fixed_string

fixed_string is a type that has been invented and re-invented repeatedly. In short, it is a class type that has the following basic representation:

template <size_t Length>
struct fixed_string {
    char _chars[Length+1] = {}; // +1 for null terminator

The “fixed” in fixed_string refers to the fixed-length nature of such a string class. A fixed_string can have a similar API to a regular string class, with an important caveat: Growing or shrinking the string generates a new type of fixed_string. That is: concatenating a fixed_string<N> with a fixed_string<M> will result in an object of type fixed_string<N + M>.

I will omit the implementation of of fixed_string’s string-like API, as it isn’t particularly interesting. I will however note this important deduction guide:

template <size_t N>
fixed_string(const char (&arr)[N])
    -> fixed_string<N-1>;  // Drop the null terminator

This allows us to do something pretty important:

fixed_string str = "Hello!";

In the above, we omit the length template argument to fixed_string, and instead rely on the deduction guide. When initialized with a string literal, the deduction guide accepting the reference-to-array will see N of 7, and thus the type of str is fixed_string<6>.

Because fixed_string<N> meets our criteria for a non-type template parameter, we can use it as such:

template <fixed_string<6> Str>
struct foo;

foo<"Hello!"> hello;
foo<"world!"> world;

Wow! Finally: We can use strings as template arguments! And the original same-ness rules still apply: the type of hello is foo<"Hello!">, and the type of world is foo<"world!">. hello and world have different types! Also:

constexpr fixed_string first = "Hel";
constexpr fixed_string second = "lo!";

foo<first + second> another;

In the above case, another has type foo<"Hello!">, exactly the same type as the hello variable in the prior example.

But there are limitations here:

foo<"nope"> b;  // FAIL!

The issue here is that "nope" is 4-char string literal, whereas foo is templatized on a 6-char fixed string. As such, the constexpr constructor for fixed_string<6> will fail, preventing us from specializing foo with anything except string literals that are 6 chars long. Pretty useless, right?

In order to fix this quirk, along with class-type non-type template parameters, C++20 allows us to use class-template argument deduction in the declaration of a template parameter:

template <fixed_string S>
struct ctad_foo {};

Here, fixed_string is the name of a class template, and not yet of a particular type. This is very weird to think about, but it combines nicely to allow us to specialize ctad_foo with any string we want:

ctad_foo<"Hello"> h
ctad_foo<"user"> u;

The type of h and u are different, and they are both using fixed_strings of different length.

NOTE: I told a small lie when I said hello has type foo<"Hello!">. It actually has type foo<fixed_string<6>{"Hello!"}>. This is actually how GCC will render the type in diagnositcs, so you should be aware. This also applies in the case of ctad_foo. The last time I checked, MSVC renders the char[] as a list of integral values, which is pretty nasty when looking at diagnostics.

Stringy Templates

Using strings as non-type template parameters opens up Pandora’s Box of possibilities. Look at this:

template <fixed_string> // [1]
struct named_type {};

template <> // [2]
struct named_type<"integer"> { using type = int; };

template <> // [2]
struct named_type<"boolean"> { using type = bool; };

template <fixed_string S> // [3]
using named_type_t = named_type<S>::type;

What’s going on here??

  1. First, we declare an empty primary definition of a class template named_type.
  2. Declare an explicit specialization of named_type for the fixed string "integer", and another for "boolean".
  3. Create an alias that grabs the nested type from a particular specialization of named_type.

Now, the madness:

named_type_t<"integer"> v = 42;
named_type_t<"boolean"> b = false;


The above code compiles and behaves exactly as one would expect, and v is of type int, while b is of type bool.

What happens if we use named_type_t with some other string?

named_type_t<"widget"> w;

This produces an error, just as you might expect:

test.cpp: In substitution of template<fixed_string<...auto...> S> using named_type_t = typename named_type::type [with fixed_string<...auto...> S = neo::fixed_string<6>{"widget"}]:
test.cpp:112:22:   required from here
test.cpp:110:7: error: no type named type in struct named_type<fixed_string<6>{"widget"}>
   110 | using named_type_t = named_type<S>::type;
       |       ^~~~~~~~~~~~

named_type_t is also SFINAE-friendly.

We can use it in even more useful contexts:

template <fixed_string Name>
concept names_a_type = requires {
    // Require that `named_type_t<Name>` produces a valid type
    typename named_type_t<Name>;


template <fixed_string S>
auto do_something() {
        "The given string must name a registered type!");

Perhaps this all seems a bit pointless, but we can take this further.

Much further.

To be continued…