Disable GC entirely

[Manu]

Sorry, void func(A* a) should be void func(A a). C++ took me for a moment ;)


On 10 April 2013 20:39, Manu <turkeyman at gmail.com> wrote:

> On 10 April 2013 20:29, Regan Heath <regan at netmail.co.nz> wrote:
>
>> Ok, lets Rewind for a sec.  I need some educating on the actual issue and
>> I want to go through the problem one case at a time..
>>
>>
>> #1 If I write a class..
>>
>> class A
>> {
>>   public int isVirt()  { return 1; }
>>   public int notVirt() { return 2; }
>> }
>>
>> and compile it with another class..
>>
>> class B : A
>> {
>>   override public int isVirt() { return 5; }
>> }
>>
>> and this main..
>>
>> void main()
>> {
>>   A a = new B();
>>   a.isVirt();    // compiler makes a virtual call
>>   a.notVirt();   // but not here
>> }
>>
>> Right?
>>
>>
>> #2 But, if I do /not/ compile class B and have..
>>
>> void main()
>> {
>>   A a = new A();
>>   a.isVirt();     // not a virtual call
>>   a.notVirt();    // neither is this
>> }
>>
>> Right?
>>
>>
>> #3 If I put class A into a library (lib/dll) then compile/link it with..
>>
>> class B : A
>> {
>>   override public int isVirt() { return 5; }
>> }
>>
>> and..
>>
>> void main()
>> {
>>   A a = new B();
>>   a.isVirt();    // compiler makes a virtual call
>>   a.notVirt();   // we're saying it has to make a virtual call here too
>> }
>>
>> So, when the compiler produced the library it compiled all methods of A
>> as virtual because it could not know if they were to be overridden in
>> consumers of the library.
>>
>> So, if the library created an A and passed it to you, all method calls
>> would have to be virtual.
>>
>> And, in your own code, if you derive B from A, then calls to A base class
>> methods will be virtual.
>>
>> Right?
>>
>> R
>>
>
> All your examples all hinge on the one case where the optimisation may
> possibly be valid, a is _allocated in the same scope_. Consider:
>
> void func(A* a)
> {
>  a.isVirt(); // is it? I guess...
>  a.notVirt(); // what about this?
> }
>
> We don't even know what a B is, or whether it even exists.
> All we know is that A's methods are virtual by default, and they could be
> overridden somewhere else. It's not possible to assume otherwise.
>
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