# What is the std::swap two-step?

This blog post has now been more than a year in the making; it’s the one for which I was laying the groundwork when I wrote “What is ADL?” (2019-04-26). Since I used the term again the other day, let’s finally define it: What is the std::swap two-step?

The name “std::swap two-step” comes (as far as I know) from Eric Niebler’s blog post “Customization Point Design in C++11 and Beyond” (October 2014). It has also been called the “std two-step” and the “ADL two-step.”

## Background

Recall that when we write an unqualified function call such as foo(x, y), the compiler will look up the name foo not only in our current namespace but also in the namespaces associated with the types of x and y. This (mis)feature of C++ is known as argument-dependent lookup, or ADL. Godbolt:

namespace N1 {
struct A {};
void foo(A&);
}
namespace N2 {
using N1::A;
A a;
void foo(A&);
}
namespace N3 {
using N1::A;
void foo(const N1::A&);

void test() {
foo(N2::a);  // calls which foo?
}
}


Here, the unqualified call foo(N2::a) calls N1::foo, because the type of N2::a (after “looking through” all using-declarations and type aliases) is N1::A. Overload resolution considers both N1::foo (found via ADL) and N3::foo (found via regular lookup) and selects N1::foo as the better-matching candidate. N2::foo is a red herring.

ADL is good for creating customization points in the same way that a sledgehammer is good for driving screws (see “Customization point design for library functions” (2018-03-19) and “C++ doesn’t know how to do customization points that aren’t operators” (2018-08-23)). When a type author wants to say “Here’s how you add two values of my type,” they do so by providing an operator+ findable by ADL. When a type author wants to say “Here’s how you print an object of my type,” they provide an operator<< findable by ADL. And when they want to say “Here’s how you swap two objects of my type,” they provide a swap(X&, X&) findable by ADL.

The canonical way to do this, in my opinion, is to use the hidden friend idiom (which is another thing I mean to blog about someday):

namespace N {
class MyType {
MyType& operator+=(const MyType&);
void print(std::ostream&);
void swap(MyType&) noexcept;

friend MyType operator+(MyType a, const MyType& b) {
a += b;
return a;
}
friend std::ostream& operator<<(std::ostream& os, const MyType& b) {
b.print(os);
return os;
}
friend void swap(MyType& a, MyType& b) noexcept {
a.swap(b);
}
};
} // namespace N


It would work just as well in this case to make those hidden friends into members of MyType’s namespace N, as follows. I never recommend this:

namespace N {
class MyType {
MyType& operator+=(const MyType&);
void print(std::ostream&);
void swap(MyType&) noexcept;
};
MyType operator+(MyType, const MyType&);
std::ostream& operator<<(std::ostream&, const MyType&);
void swap(MyType&, MyType&) noexcept;
}


Anyway, the upshot of this is that when some client code — some generic code, perhaps — makes an unqualified function call to one of these names, they’ll trigger ADL and ADL will find these candidates.

void test(N::MyType& t) {
t + t;           // considers N::operator+
std::cout << t;  // considers N::operator<<
swap(t, t);      // considers N::swap
}


## What about types that haven’t customized their behavior?

What happens if we have a type like struct S that hasn’t customized its behavior for addition, printing, or swapping?

struct S {};
void test(S& s) {
s + s;           // ERROR!
std::cout << s;  // ERROR!
swap(s, s);      // ERROR!
}


In the first two cases, it makes sense: You can’t add instances of a class type that doesn’t explicitly overload +, and you can’t print instances of a class type that doesn’t explicitly overload <<. But (I claim) it doesn’t really make sense to say that you can’t swap instances.

There are some other operations that C++ lets you do on “vanilla” class types like S: default-construction, copy-construction, copy-assignment, and destruction. C++ handles those operations by generating an implicitly defaulted definition for each of those operations. C++ could have said that it’d also generate an implicitly defaulted definition for swap; but history didn’t go that way.

Instead, the standard library provides a sort of “hand-written default implementation” for swap, and leaves it up to the client code to glue the pieces together correctly. In generic code, where you don’t know whether your type T has an ADL candidate for swap or not, you want basically this high-level behavior:

template<class T>
void test(T& t) {
if (t has ADL swap) {
swap(t, t);  // using the ADL candidate
} else {
std::swap(t, t);  // using the library's hand-written default
}
}


It turns out that we can achieve this behavior very easily, because the STL’s std::swap is deliberately written as the least-constrained kind of template possible, meaning that it ranks lowest in the overload resolution scheme. Any viable candidate found via ADL ought to rank higher than std::swap. So in real life, we just need to make sure that name lookup will always consider std::swap a candidate in addition to the ADL candidate(s) (if any). We write this:

template<class T>
void test(T& t) {
using std::swap;  // make it a candidate
swap(t, t);
}


And here we finally have the “two-step.”

• Step 1: using std::swap;

• Step 2: Unqualified call to swap(t, t).

## When should I use the two-step?

The std::swap two-step solves a problem in static dispatch that’s very similar to a problem you might already know from dynamic dispatch.

class Base {
void frotz() { puts("Light"); }
virtual void xyzzy() { puts("Debris"); }
virtual void plugh() = 0;
};

class Derived : public Base { ... };

void test(Base *b) {
b->frotz();
b->xyzzy();
b->plugh();
}


In C++, where static dispatch is the status quo, an ordinary non-virtual method like frotz is basically the type author telling the program, “I know exactly how to frotz myself. This is how you do it. Nobody else will ever know better than me how to frotz.”

A virtual method like xyzzy, in contrast, indicates, “I know how to xyzzy myself, but my child class might know a better way. If my child has an opinion, you should trust my child over me.”

A pure virtual method like plugh indicates, “Not only should you trust my child over me, but I’m not giving you a choice! I have no idea how to plugh myself. My child must come up with a solution. I offer no guidance here.”

Translating this into the world of static dispatch and ADL, we have something like this:

namespace base {
template<class T> void frotz(T t) { puts("Light"); }
template<class T> void xyzzy(T t) { puts("Debris"); }
}

template<class T>
void test(T t) {
base::frotz(t);
using base::xyzzy; xyzzy(t);
plugh(t);
}


A qualified call like base::frotz(t) indicates, “I’m sure I know how to frotz whatever this thing may be. No type T will ever know better than me how to frotz.”

An unqualified call using the two-step, like xyzzy(t), indicates, “I know how to xyzzy whatever this thing may be, but type T might know a better way. If T has an opinion, you should trust T over me.”

An unqualified call not using the two-step, like plugh(t), indicates, “Not only should you trust T over me, but I myself have no idea how to plugh anything. Type T must come up with a solution; I offer no guidance here.”

Notice that in the virtual world, class Base itself has control over which calls dispatch to the child and which don’t. In the static-dispatch world, it’s the client code, test, that decides what kind of call to use in each situation.

### Should I use the two-step in non-generic code?

No. Sometimes I see students want to do this:

class Book {
std::string title_;
int pagecount_;
public:
void swap(Book& rhs) noexcept {
using std::swap;
swap(title_, rhs.title_);
swap(pagecount_, rhs.pagecount_);
}
};


All else being equal, the two-step here is just obfuscation. We know that title_ is a std::string; we know what candidate we want to use to swap strings. We also know what candidate we want to use to swap ints. I would prefer to see this written simply as

    void swap(Book& rhs) noexcept {
std::swap(title_, rhs.title_);
std::swap(pagecount_, rhs.pagecount_);
}


or even

    void swap(Book& rhs) noexcept {
title_.swap(rhs.title_);
std::swap(pagecount_, rhs.pagecount_);
}


(Note that std::swap is overloaded for strings, so std::swap(s1, s2) and s1.swap(s2) end up doing the exact same thing. The latter is no more efficient than the former; it’s just shorter to type.)

### Should I use the two-step to provide a fallback in generic code?

Yes. The “std::swap two-step” gets its name from exactly this use-case. The STL provides a fallback implementation of swap that we want to use specifically for types T that haven’t provided any better candidate.

You should use the two-step in any similar situation. In my experience, these situations are very few and far between. Theoretically, you should use the two-step on customization points like begin and end:

template<class R>
void mysort(R& range) {
using std::begin;
using std::end;
std::sort( begin(range), end(range) );
}


This means that mysort<N::MyType> will consider using N::begin and N::end if they exist and are better matches than the generic std::begin and std::end. However, in my experience, this is never actually the case. Lots of type authors provide an ADL swap. No type author has ever provided an ADL begin.

However, should you use std::begin(x) instead of x.begin()? In generic code, yes; because std::begin(x) works for array types like int[10] whereas x.begin() works only for class types.

UPDATE, 2020-07-12: Reddit points out that std::filesystem::directory_iterator is a type with an ADL begin and no member begin. I should have remembered this because I wrote the rejected proposal P0757 “regex_iterator should be iterable” (September 2017)! Michael Hava reports that Microsoft’s C++/CX also provides an ADL begin taking an IVector<T>^ (both in the Windows::Foundation::Collections namespace), thus enabling you to loop for (auto&& elt : myVec) without dereferencing myVec first. These are good examples of real types providing ADL begin in lieu of member begin. “Good example” does not mean “good role model,” though!

### Should I use the two-step in generic code lacking a fallback?

No. See the end of “How to erase from an STL container” (2020-07-08), where I say that it would be wrong wrong wrong to do this:

template<class C>
void remove_all_odd_elements(C& c) {
using std::erase_if;
erase_if(c, [](int x) { return x % 2; });
}


There are two possible readings here: Either the author of this generic code intends erase_if to be like frotz — “No type T will ever know better than me how to frotz things” — or else the author intends it to be like xyzzy — “I know how to xyzzy things, but if T has an opinion, you should trust T over me.”

In the former case, bringing ADL into the mix is simply incorrect. The author should have written:

template<class C>
void remove_all_odd_elements(C& c) {
std::erase_if(c, [](int x) { return x % 2; });
}


In the latter case, the author is permitting C to have a say in the meaning of erase_if; but also (by using the two-step) saying that it’s okay to fall back to std::erase_if if C doesn’t provide any viable erase_if via ADL. The problem with this is that the erase_if in namespace std works only for container types that themselves are already in namespace std! There is no situation where std::erase_if would ever actually work as a “fallback.” Therefore, the author should simply have written:

template<class C>
void remove_all_odd_elements(C& c) {

// use ADL so we'll consider
// MyLib::erase_if(MyLib::FlatMap&, Predicate)
// as a candidate for this call

erase_if(c, [](int x) { return x % 2; });
}


This is analogous to how you wouldn’t write using std::operator+; in front of a call to c + c. There’s simply no situation in which the ADL call would have failed and yet some overload of std::operator+ would have worked.

However, as I said in that previous blog post, I really don’t think this is a good idea in the specific case of erase_if. Call std::erase_if with qualification, the same way you’d call std::rethrow_exception or std::from_chars.

## Can’t I just use C++20 std::ranges::swap?

Yes, my understanding is that in C++20 you can just write

#include <concepts>

template<class T>
void test(T& t) {
std::ranges::swap(t, t);
}


std::ranges::swap is not a function but a CPO, i.e., a global variable with a templated operator() that has been defined to do the two-step internally so that you don’t have to.

But for those who need portability to C++17-and-earlier; or for those who might encounter those rare non-swap customization points; or for those who just like to understand what std::ranges::swap is doing (and why!); I hope this blog post helped.

Posted 2020-07-11