Why pure virtual function is initialized by 0?

Question

We always declare a pure virtual function as :

virtual void fun () = 0 ;

i.e. it is always assigned to 0.

What I understand is that this is to initialize the vtable entry for this function to NULL and any other value here results in a compile time error. Please tell me whether this understanding is correct or not.

Answer 1

This deliberately creates an abstract virtual, which makes the class an abstract class.

It forces any programmers who use it to declare a subclass to implement the missing features.

This might be used where a common base class is simply there to implement some common features, but does not have enough implementation to do anything useful itself.

Answer 2

I'm not sure if there is any meaning behind this, it is just the syntax of the language.

Answer 3

I would assume that this is just part of the C++ grammar. I don't think there are any restrictions to how the compilers actually implement this for a given specific binary format. You're assumption probably was right for the early day C++ compilers.

Answer 4

C++ must have a way to distinguish a pure virtual function from a declaration of a normal virtual function. They chose to use the = 0 syntax. They could just have easily done the same by adding a pure keyword. But C++ is pretty loath to add new keywords and prefers to use other mechanisms to introduce features.

Answer 5

Nothing is "initilaized" or "assigned" zero in this case. = 0 in just a syntactical construct consisting of = and 0 tokens, which has absolutely no relation to either initialization or assignment.

It has no relation to any actual value in "vtable". The C++ language has no notion of "vtable" or anythng like that. Various "vtables" are nothing more than just details of specific implementations.

Answer 6

The reason =0 is used is that Bjarne Stroustrup didn't think he could get another keyword, such as "pure" past the C++ community at the time the feature was being implemented. This is described in his book, The Design & Evolution of C++, section 13.2.3:

The curious =0 syntax was chosen ... because at the time I saw no chance of getting a new keyword accepted.

He also states explicitly that this need not set the vtable entry to NULL, and that doing so is not the best way of implementing pure virtual functions.

Answer 7

I remember reading that the justification for the funny syntax was that it was easier (in terms of standards-acceptance) than introducing another keyword that would do the same thing.

I believe this was mentioned in The Design and Evolution of C++ by Bjarne Stroustrup.

Answer 8

The = 0declares a pure virtual function.

What is understand is that this is to initialize the vtable entry for this function to NULL and any other value here results in compile time error

I don't think that's true. It's just special syntax. The vtable is implementation-defined. No one says a vtable entry for a pure member must be actually zeroed upon construction (although most compilers handle vtables similar).

Answer 9

Well, you can also initialize the vtable entry to point to an actual function"

 virtual void fun()
 {
     //dostuff()
 }

Seems intuitive that the vtable entry can either be defined to point nowhere (0) or to a function. Letting you specify your own value for it would probably result in it pointing to garbage instead of to a function. But that is why "= 0" is allowed and "= 1" is not. I suspect Neil Butterworth is right about why "= 0" is used at all

Answer 10

Section 9.2 of the C++ standard gives the syntax for class members. It includes this production:

pure-specifier:
    = 0

There is nothing special about the value. "= 0" is just the syntax for saying "this function is pure virtual."

Answer 11

It makes the function a pure virtual function - one without a default implementation. A class with a pure virtual cannot be instantiated, it must be sub-classed.

Answer 12

C++ has always shied away from introducing new keywords, since new reserved words break old programs which use these words for identifiers. It's often seen of one of the language's strengths that it respects old code as far as possible.

The =0 syntax might indeed have been chosen since it resembles setting a vtable entry to 0, but this is purely symbolic. (Most compilers assign such vtable entries to a stub which emits an error before aborting the program.) The syntax was mainly chosen because it wasn't used for anything before and it saved introducing a new keyword.

Answer 13

As with most "Why" questions about the design of C++, the first place to look is The Design and Evolution of C++, by Bjarne Stroustrup¹:

The curious =0 syntax was chosen over the obvious alternative of introducing a new keyword pure or abstract because at the time I saw no chance of getting a new keyword accepted. Had I suggested pure, Release 2.0 would have shipped without abstract classes. Given a choice between a nicer syntax and abstract classes, I chose abstract classes. Rather than risking delay and incurring the certain fights over pure, I used the tradition C and C++ convention of using 0 to represent "not there." The =0 syntax fits with my view that a function body is the initializer for a function also with the (simplistic, but usually adequate) view of the set of virtual functions being implemented as a vector of function pointers. [ ... ]

¹§13.2.3 Syntax