pure-specifier on function-definition

Question

While compiling on GCC I get the error: pure-specifier on function-definition, but not when I compile the same code using VS2005.

class Dummy {   
  //error: pure-specifier on function-definition, VS2005 compiles 
  virtual void Process() = 0 {};
};

But when the definition of this pure virtual function is not inline, it works:

class Dummy
{
  virtual void Process() = 0;
};
void Dummy::Process()
{} //compiles on both GCC and VS2005

What does the error means? Why cannot I do it inline? Is it legal to evade the compile issue as shown in the second code sample?

By the way you've got a superfluous semicolon. – QuentinUK Aug 13 '22 at 18:00 — QuentinUK, Aug 13 '22 at 18:00

Dima · Accepted Answer · 2010-06-01T16:10:46.557

40

Ok, I've just learned something. A pure virtual function must be declared as follows:


class Abstract 
{
public:
   virtual void pure_virtual() = 0;
};

It may have a body, although it is illegal to include it at the point of declaration. This means that to have a body the pure virtual function must be defined outside the class. Note that even if it has a body, the function must still be overridden by any concrete classes derived from Abstract. They would just have an option to call Abstract::pure_virtual() explicitly if they need to.

The details are here.

edited Jun 01 '10 at 16:10

answered Jun 01 '10 at 15:58

Dima

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Yes it should, if you need one. It's perfectly legal to have one. – Jun 01 '10 at 16:00
But with a body it would just be a virtual function. What should a pure virtual function with a body do??? – Martin Tilsted Jun 01 '10 at 16:04
1

@Martin Possibly nothing - if you declare a pure virtual destructor (for example) you must give it a body. – Jun 01 '10 at 16:06
@MartinTilsted Not really, see http://stackoverflow.com/questions/5481941/c-pure-virtual-function-have-body. – Melebius Jan 21 '14 at 08:30

score 20 · Answer 2 · answered Jun 01 '10 at 16:03

20

C++ Standard, 10.4/2:

a function declaration cannot provide both a pure-specifier and a definition

answered Jun 01 '10 at 16:03

Paul

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This is not normative but taken from a note. – Columbo Apr 19 '15 at 19:28
@Columbo Well, according to §6.5.1 of ISO/IEC Directives Part 2 notes shall not contain permissions at all. While this note clearly contains one, which mislead me. Still modern GCC and Clang don't allow this, do they violate the standard? – Paul Apr 20 '15 at 12:34
Protip: Read my answer. The note *is* correct, but to 'prove' this one has to show up the grammar rules. – Columbo Apr 20 '15 at 12:37

score 13 · Answer 3 · answered Jun 01 '10 at 16:00

13

This syntax:

virtual void Process() = 0 {};

is not legal C++, but is supported by VC++. Exactly why the Standard disallows this has never been obvious to me. Your second example is legal.

answered Jun 01 '10 at 16:00

Amardeep AC9MF · Answer 4 · 2010-06-01T16:09:18.717

4

Pure virtual functions in C++ by definition have no definition in the declaration.

You second code block is not avoiding the compiler issue. It is implementing a pure virtual function the way it was intended.

The question to ask is, why do you need to declare it pure virtual if you intend to have a default implementation?

edited Jun 01 '10 at 16:09

answered Jun 01 '10 at 16:01

Amardeep AC9MF

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Pure virtual functions may have definition. Think about pure virtual destructor - it MUST be defined. – Paul Jun 01 '10 at 16:03
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Please enlighten me with the purpose of a pure virtual destructor? – Amardeep AC9MF Jun 01 '10 at 16:06
1

You make a destructor pure when you don't want its class instantiated and there are no other PVF candidates. Quite a rare requirement, I agree. – Jun 01 '10 at 16:10
3

> why do you need to declare it pure virtual if you intend to have a default implementation? As an author of the code in question I will answer: the default implementation is provided as a helper for derived classes. They are expected to call the default implementation, but they need to do so explicitly. This is to make sure calling default implementation is a conscious design, and not an unintended behaviour. Kind like "explicit" with conversion / construction. It does not provide any new possibilities, but it can help avoiding mistakes. – Suma Jun 01 '10 at 17:02

score 3 · Answer 5 · answered Apr 19 '15 at 19:50

This is gramatically disallowed - the declarator that can include pure-specifiers, i.e. the member-declarator, only appears in declarations that aren't definitions. [class.mem] :

member-declaration:
         attribute-specifier-seq_opt decl-specifier-seq_opt member-declarator-list_opt ;
         function-definition
         [...]

member-declarator-list:
         member-declarator
         member-declarator-list , member-declarator

member-declarator:
         declarator virt-specifier-seq_opt pure-specifier_opt
         declarator brace-or-equal-initializer_opt
         identifier_opt attribute-specifier-seq_opt : constant-expression

The grammar of function-definition does not include a pure-specifier, [dcl.fct.def.general]:

function-definition:
attribute-specifier-seq_opt decl-specifier-seq_opt declarator virt-specifier-seq_opt function-body

Nikolai Fetissov · Answer 6 · 2010-06-01T16:20:22.917

0

You can certainly provide a body for pure virtual function. That function will be pointed to by that abstract class vtable. Otherwise the same slot will point to compiler-specific trap function like __cxa_pure_virtual for GCC. There's of course nothing about this in the standard.

edited Jun 01 '10 at 16:20

answered Jun 01 '10 at 16:09

Nikolai Fetissov

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pure-specifier on function-definition

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