Will std::swap still be defined by including algorithm in C++0x?

Question

The swap function template was moved from <algorithm> to <utility> in C++0x. Does the former include the latter in C++0x? Or do they both include a common header the defines swap?

In other words, is the following code guaranteed to compile in C++0x?

#include <algorithm>   // will this pull in std::swap?

// ...

using std::swap;
swap(a, b);

Answer 1

The FDIS (n3290), in Annex C, "Compatibility", C.2.7 says:

17.6.3.2

Effect on original feature: Function swap moved to a different header

Rationale: Remove dependency on <algorithm> for swap.

Effect on original feature: Valid C++ 2003 code that has been compiled expecting swap to be in <algorithm> may have to instead include <utility>.

So no, it's not guaranteed to compile, this is intentionally a breaking change. Whether individual implementations will actually break C++03 code is another matter. As you point out it's easy enough for them not to, by defining swap via either header. But there's a choice between making it easier to port C++03 code to C++0x, vs. helping people write strictly conforming C++0x.