Two kinds of function template parameters

When you’re writing a function template in C++, its template parameters will naturally fall into two mutually exclusive categories:

  • Template parameters that the caller must supply in angle brackets;
  • Template parameters that the caller must not supply in angle brackets.

The core language doesn’t distinguish between these two kinds of template parameters; they go into the same template parameter list. The only real restriction is that all parameters of the first kind must necessarily precede all parameters of the second kind.

The poster child for this is std::make_unique.

template<class T, class... Args>
std::unique_ptr<T> make_unique(Args&&...);

The caller must supply T in their angle brackets. The caller must not supply any of Args... in their angle brackets. Again, that “must not” is a semantic rule; the language doesn’t physically enforce it. You physically can call make_unique like this:

auto p = std::make_unique<std::string, int>(3, 'c'); // Wrong

But that’s not how make_unique is intended to be used. It’s supposed to infer Args... from the types of its function arguments.

In fact, the STL reserves the right to break callers who pass template arguments they’re not supposed to. For example, back in C++98 (post-LWG181) the signature of make_pair was:

template<class T, class U>
std::pair<T, U> make_pair(T, U);

and so you could physically write:

int intval = 42;
auto p1 = std::make_pair<int, std::string>(42, "abc");
auto p2 = std::make_pair<int, std::string>(intval, "abc");

But in C++11, the signature of make_pair changed to use forwarding references instead of const&:

template<class T, class U>
std::pair<T, U> make_pair(T&&, U&&);

The initializer of p1 above continues to compile (because int&& can bind to 42), but p2 now fails to compile (because int&& won’t bind to the lvalue intval; T won’t deduce as int& because we explicitly supplied it as int in the angle brackets in exactly the way I’m telling you not to).

Lots more examples exist; many are so trivial that you wouldn’t even recognize them as examples. E.g.:

int a[100];
std::sort(a, a+100); // Right

We physically could write this call as:

std::sort<int*>(a, a+100); // Wrong

But we shouldn’t, because sort isn’t intended to be called with angle brackets. By writing the angle brackets, not only are we obfuscating our own code — we’re also risking breakage when we upgrade our vendor’s STL.

I believe we’re formally guaranteed that a conforming implementation of C++23 will support calling std::sort<int*> with those template parameters, because [alg.sort] implies as much. But we’re not guaranteed that the wording of [alg.sort] won’t change! A future revision of C++ might change sort’s template parameters, in the same way C++11 changed make_pair’s template parameters.

UPDATE, 2024-01-31: No, in fact [algorithm.requirements]/15 already makes it “unspecified” both whether std::sort<int*> is well-formed and (if so) what it does. The “Wrong” line is, officially, wrong.

Another classic example is swap (with or without the std::swap two-step):

Widget a, b;
Widget wa[10], wb[10];

using std::swap;

swap<Widget>(a, b); // Wrong
swap(a, b); // Right

swap<Widget>(wa, wb); // Wrong; compiles
swap<Widget[10]>(wa, wb); // Wrong; doesn't compile
swap(wa, wb); // Right

std::string sa, sb;
swap<std::string>(sa, sb); // Wrong; inefficient
swap(sa, sb); // Right; efficient

Finally, the most prominent case where you’ll be tempted to use the angle brackets, but I’m telling you it’s better not to: min and max.

int i = std::min<int>(strlen(s), 42);    // Bad

int i = std::min(int(strlen(s)), 42);    // Better
int i = std::min(strlen(s), size_t(42)); // Better
int i = std::min(strlen(s), 42uz);       // Better (C++23)

[algorithm.requirements]/15 doesn’t apply here because [alg.min.max]/4 overrules it. Here, you really do have to take my word for it that the “Bad” line is bad.

What about “maybe-supply” function template parameters?

A class template like vector can be instantiated as either vector<int> or vector<int, Alloc>. That allocator parameter sure looks like it doesn’t fall into either of my categories — it’s a “maybe-supply” template parameter! But notice that vector isn’t a function template. Non-function templates obey a completely different model from function templates:

  • Non-function templates don’t allow you to pass only some template arguments in their angle brackets; e.g. std::tuple<int, int>{1, 2, 3} doesn’t compile. You can omit a trailing template parameter only if the type author has provided a default for that parameter, which indicates they’re explicitly thinking about that situation.

  • Since C++17, class templates can omit the entire angle-bracket part (this is called CTAD and I recommend never to use it). But most class templates, e.g. std::vector, aren’t designed to play well with CTAD; they require you to specify at least one argument, and as soon as you write the angle brackets, you’re physically required to spell out all the (non-defaulted) template arguments.

Still, what if I want to design a function that can take or not take an explicitly supplied template argument? Something like make_optional, which can be called as either:

auto o1 = make_optional("abc");         // optional<const char*>
auto o2 = make_optional<string>("abc"); // optional<string>

(This is a bad idea.) Now, what we want to write would simply deduce the function argument’s type A, and then we’d default the “maybe-supply” template argument T to A:

template<class A, class T = A>
optional<T> make_optional(A);

But this places A in front of T, which means the caller must supply A in order to supply T! And we can’t switch A with T because then A would be used before it was declared. An ugly, but viable, solution is (Godbolt):

template<class T = void, class A, class R = conditional_t<is_void_v<T>, A, T>>
optional<R> make_optional(A);

A cleaner solution is simply to provide two overloads, as std::make_optional does in real life (Godbolt):

template<class T, class A>
optional<T> make_optional(A);
template<class A>
optional<A> make_optional(A);

But then you should ask why you’re putting these two functions in an overload set in the first place. (See “Inheritance is for sharing an interface (and so is overloading)” (2020-10-09).) What do you gain by naming them make_optional(x) and make_optional<T>(x), as opposed to, let’s say, make_optional(x) and optional<T>(in_place, x)? I claim: Very little. (See “API design advice from Anakin and Obi-Wan” (2022-12-17).) The game isn’t worth the candle.

Overload sets should never take “maybe-supply” template parameters.

How to enforce a “must-supply” parameter

Very rarely, you might write a template where a “must-supply” parameter is also deducible from the function argument list.

template<class T>
T implicitly_convert_to(T t) {
  return t;

// Intended usage:
auto i = implicitly_convert_to<int>(3.14);

// Preventable mistake:
auto j = implicitly_convert_to(3.14);

In this case, if we accidentally omitted <int>, the call would still compile: T would simply be deduced as double, and we’d get unexpected behavior at compile time. You can prevent this from compiling — and force the caller to pass T explicitly in the angle brackets — by applying a “deduction firewall” to function parameter t. In C++20, that’s spelled type_identity<T>::type, or just type_identity_t<T> for short.

template<class T>
T implicitly_convert_to(std::type_identity_t<T> t) {
  return t;

auto i = implicitly_convert_to<int>(3.14); // Right, compiles
auto j = implicitly_convert_to(3.14);      // Wrong, correctly doesn't compile

How to enforce a “must-not-supply” parameter

More commonly, you’ll write a template involving “must-not-supply” parameters, and (out of sheer paranoia) you’ll want to guard against the caller’s accidentally supplying a value for one of them.

template<class T, class A>
T implicitly_convert_to(A a) {
  return a;

// Intended usage: i is 9
auto i = implicitly_convert_to<int>(9.9999999);

// Preventable mistake: j is 10
auto j = implicitly_convert_to<int, float>(9.9999999);

The basic idea is to insert a “stack canary” between the must-supply and must-not-supply parameters, like this:

template<class T, class Canary = void, class A>
T implicitly_convert_to(A a) {
  static_assert(is_void_v<Canary>, "Only one explicit template parameter, please");
  return a;

// No longer compiles
auto j = implicitly_convert_to<int, float>(9.9999999);

// But a truly hostile caller can still do it: k is 10
auto k = implicitly_convert_to<int, void, float>(9.9999999);

An anonymous reader points out that you can make this truly foolproof — and eliminate the <type_traits> dependency at the same time — by making the canary a pack. At least in present-day C++, the first pack inexorably expands to consume all the remaining explicit template arguments, so our hostile caller can’t bypass it.

So instead of a “canary,” it’s more of a “pelican.” A pack also acts as an explicit-argument-specification firewall; but I’ve called one thing a “firewall” in this blog post already. To stretch a plumbing metaphor, we might call it a “surge tank”: it goes unused in normal operation, but when too many arguments come in at once, it can safely hold the excess so as not to damage the infrastructure downstream of it.

template<class T, class... Canary, class A>
T implicitly_convert_to(A a) {
  static_assert(sizeof...(Canary) == 0, "Only one explicit template parameter, please");
  return a;

// No longer compile
auto j = implicitly_convert_to<int, float>(9.9999999);
auto k = implicitly_convert_to<int, void, float>(9.9999999);

A third kind of template parameter

Pre-C++20, we had a third kind of template parameter too; the kind that you must not supply in the angle brackets, but which wouldn’t be used in the template’s body either. This kind of parameter encodes what we now call “constraints” on the template.

template<class T, class A,
         class E = enable_if_t<is_convertible_v<A, T>>>
T implicitly_convert_to(A arg) { return T(arg); }

Here T is “must-supply,” A is “must-not-supply,” and E is a “constraint.” convert won’t participate in overload resolution unless is_convertible_v<T, A> is true… or, unless the caller breaks our rule and supplies E in the angle brackets!

int i = 42;
auto j = implicitly_convert_to<intptr_t>(&i); // Good; doesn't compile
auto k = implicitly_convert_to<intptr_t, int*, void>(&i); // Bad; compiles!

Pre-C++20, you can make it foolproof by using a non-type template parameter:

template<class T, class A,
         enable_if_t<is_convertible_v<A, T>, int> E = 0>
T implicitly_convert_to(A arg) { return T(arg); }

In C++20, this kind of template parameter should eventually go extinct, because we can always replace it with a requires clause instead. See “Prefer core-language features over library facilities” (2022-10-16).

template<class T, class A>
  requires is_convertible_v<A, T>
T implicitly_convert_to(A arg) { return T(arg); }
Posted 2023-12-17