Is std::vector<T> a `user-defined type`?

40
votes

In 17.6.4.2.1/1 and 17.6.4.2.1/2 of the current draft standard restrictions are placed on specializations injected by users into namespace std.

The behavior of a C ++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified. A program may add a template specialization for any standard library template to namespace std only if the declaration depends on a user-defined type and the specialization meets the standard library requirements for the original template and is not explicitly prohibited.

I cannot find where in the standard the phrase user-defined type is defined.

One option I have heard claimed is that a type that is not std::is_fundamental is a user-defined type, in which case std::vector<int> would be a user-defined type.

An alternative answer would be that a user-defined type is a type that a user defines. As users do not define std::vector<int>, and std::vector<int> is not dependent on any type a user defines, std::vector<int> is not a user-defined type.

A practical problem this impacts is "can you inject a specialization for std::hash for std::tuple<Ts...> into namespace std? Being able to do so is somewhat convenient -- the alternative is to create another namespace where we recursively build our hash for std::tuple (and possibly other types in std that do not have hash support), and if and only if we fail to find a hash in that namespace do we fall back on std.

However, if this is legal, then if and when the standard adds a hash specialization for std::tuple to namespace std, code that specialized it already would be broken, creating a reason not to add such specializations in the future.

While I am talking about std::vector<int> as a concrete example, I am trying to ask if types defined in std are ever user-defined type s. A secondary question is, even if not, maybe std::tuple<int> becomes a user-defined type when used by a user (this gets slippery: what then happens if something inside std defines std::tuple<int>, and you partial-specialize hash for std::tuple<Ts...>).

There is currently an open defect on this problem.

4
c++c++11language-lawyertemplate-specializationc++14
It doesn't mean a type that's defined by the programmer? A type that's not already present in the C++ specification? – Robert Harvey
@Robert No, at least not historically. Historically, a UDT is any non-fundamental type, i.e. any type other than enums and builtin types such as int, char etc. But I agree that this makes the usage in the context cited here weird. – Konrad Rudolph
Yes, that's essentially my understanding. Anything you might define with class or struct. – Robert Harvey
@Robert Forget it, brain fart. I was thinking of (non-)POD, not UDT. – Konrad Rudolph
BTW: I've seen this type of question several times and I always wondered why the definition of "user-defined type" is questioned instead of simply questioning if the restrictions in 17.6.4.2.1 simply need to be fixed (by extending the wording to not just rely on UDTs) – Daniel Frey

4 Answers

16
votes

When Clause 17 says "user-defined" it means "a type not defined in the standard" so std::vector<int> is not user-defined, neither is std::string, so you cannot specialize std::vector<int> or std::vector<std::string>. On the other hand, struct MyClass is user-defined, because it's not a type defined in the standard, so you can specialize std::vector<MyClass>.

This is not the same meaning of "user-defined" used in clauses 1-16, and that difference is confusing and silly. There is a defect report for this, with some discussion recorded that basically says "yes, the library uses the wrong term, but we don't have a better one".

So the answer to your question is "it depends". If you're talking to a C++ compiler implementor or a core language expert, std::vector<int> is definitely a user-defined type, but if you're talking to a standard library implementor, it is not. More precisely, it's not user-defined for the purposes of 17,6.4.2.1.

One way to look at it is that the standard library is "user code" as far as the core language is concerned. But the standard library has a different idea of "users" and considers itself to be part of the implementation, and only things that aren't part of the library are "user-defined".

Edit: I have proposed changing the library Clauses to use a new term, "program-defined", which means something defined in your program (as opposed to UDTs defined in the standard, such as std::string).

33
votes

Prof. Stroustrup is very clear that any type that is not built-in is user-defined. See the second paragraph of section 9.1 in Programming Principles and Practice Using C++.

He even specifically calls out “standard library types” as an example of user-defined types. In other words, a user-defined type is any compound type.

Source

The article explicitly mentions that not everyone seems to agree, but this is IMHO mostly wishful thinking and not what the standard (and Prof. Stroustrup) are actually saying, only what some people want to read into it.

7
votes

As users do not define std::vector<int>, and std::vector<int> is not dependent on any type a user defines, std::vector<int> is not a user-defined type.

The logical counter argument is that users do define std::vector<int>. You see std::vector is a class template and as such has no direct representation in binary code.

In a sense it gets it binary representation through the instantiation of a type, so the very action of declaring a std::vector<int> object is what gives "soul" to the template (pardon the phrasing). In a program where noone uses a std::vector<int> this data type does not exist.

On the other hand, following the same argument, std::vector<T> is not a user defined type, it is not even a type, it does not exist; only if we want to (instantiate a type), it will mandate how a structure will be layed out but until then we can only argue about it in terms of structure, design, properties and so on.

Note

The above argument (about templates being not code but ... well templates for code) may seem a bit superficial but draws it's logic, from Mayer's introduction in A. Alexandrescu's book Modern C++ Design. The relative quote there, goes like this :

Eventually, Andrei turned his attention to the development of template-based implementations of popular language idioms and design patterns, especially the GoF[*] patterns. This led to a brief skirmish with the Patterns community, because one of their fundamental tenets is that patterns cannot be represented in code. Once it became clear that Andrei was automating the generation of pattern implementations rather than trying to encode patterns themselves, that objection was removed, and I was pleased to see Andrei and one of the GoF (John Vlissides) collaborate on two columns in the C++ Report focusing on Andrei's work.

2
votes

The draft standard contrasts fundamental types with user-defined types in a couple of (non-normative) places.

The draft standard also uses the term "user-defined" in other contexts, referring to entities created by the programmer or defined in the standard library. Examples include user-defined constructor, user-defined operator and user-defined conversion.

These facts allow us, absent other evidence, to tentatively assume that the intent of the standard is that user-defined type should mean compound type, according to historical usage. Only an explicit clarification in a future standard document can definitely resolve the issue.

Note that the historical usage is not clear on types like int* or struct foo* or void(*)(struct foo****). They are compound, but should they (or some of them) be considered user-defined?