On 20 January 2023 07:32:56 GMT, Bo Persson via Std-Proposals <std-proposals_at_[hidden]> wrote:
>On 2023-01-20 at 07:55, Julien Allali via Std-Proposals wrote:
>> Hi everyone,
>>
>> I checked the following code from one of my student:
>>
>> class I{
>> public:
>> virtual ~I(){}
>> virtual void print(){
>> printf("hello %p\n",this);
>> }
>> virtual void f()=0;
>> };
>>
>> class B : public virtual I{
>> int i,j;
>> public:
>> ~B(){}
>> virtual void f(){
>> printf("B %p\n",this);
>> }
>> };
>>
>>
>> void f(I *i){
>> for(int j=0;j<5;++j)
>> {
>> i->print();
>> i->f();
>> printf("i=%p ",i);
>> i++; // PROBLEM IS HERE
>> printf(" => %p \n",i);
>> }
>> }
>>
>> int main(){
>> B b[]={B(),B()};
>> printf("B: %p %p\n",b,b+1);
>> f(b);
>> }
>>
>> As you can see, I is a non instantiable type as it has one pure virtual method. My concern is about the line "i++": I believed it will lead to an error or at least a warning (g++ 11.3.0)... Indeed, I can not imagine a valid case where doing arithmetic on a pointer of a type with pure virtual function can be valid.... Shouldn't the standard forbid such arithmetic?
>>
>
>Incrementing a pointer to a single element creates a one-past-the-end pointer for the type. That is ok to *have*, but of course you cannot dereference it.
>
>
>

Why do even abstract base classes have a well-formed sizeof?

https://www.eel.is/c++draft/expr.sizeof#1.sentence-1

"The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand."

An abstract base class can never be non-overlapping, so the value has no meaning.

If an abstract base did not have well-formed sizeof, then not having well-formed pointer arithmetic would also be consistent with this.

Cheers,
Lénárd