@Naman : Can you tell me why the value of dval is accessed via bval of base
class ? I got ur point but still confused about it.
Sanju
:)
On Sun, Aug 21, 2011 at 12:16 AM, Naman Mahor <naman.ma...@gmail.com> wrote:
> #include<iostream>
> using namespace std;
>
> class base
> {
> public:
> int bval;
> base(){bval=0;}
> };
> class deri:public base
> {
> public:
> int dval;
> deri(){dval=1;}
> };
> void SomeFunc(base *arr,int size)
> {
> for(int i=0; i<size; i++,arr++)
> { cout<<arr->bval<<endl;
> cout<<&(arr->bval)<<endl;
>
>
> }
> cout<<endl;
> }
> int main()
> {
> base BaseArr[5];
> SomeFunc(BaseArr,5);
> deri DeriArr[5];
> SomeFunc(DeriArr,5);
> system("pause");
> return 0;
> }
> see the output u will understant
>
>
> On Sun, Aug 21, 2011 at 12:44 PM, Puneet Chawla <puneetchawla...@gmail.com
> > wrote:
>
>> Wil u please explain in context of o/p.means how 1 whn
>> SomeFunc(DeriArr,5); stmt is executed how 0 and 1 will come..???
>>
>>
>> On Sun, Aug 21, 2011 at 12:40 PM, Naman Mahor <naman.ma...@gmail.com>wrote:
>>
>>> because the pointer of base class will move by 2 bytes every time but the
>>> size of derived class is 4 byte.
>>>
>>>
>>> On Sun, Aug 21, 2011 at 12:31 PM, Puneet Chawla <
>>> puneetchawla...@gmail.com> wrote:
>>>
>>>> class base
>>>> {
>>>> public:
>>>> int bval;
>>>> base(){bval=0;}
>>>> };
>>>> class deri:public base
>>>> {
>>>> public:
>>>> int dval;
>>>> deri(){dval=1;}
>>>> };
>>>> void SomeFunc(base *arr,int size)
>>>> {
>>>> for(int i=0; i<size; i++,arr++)
>>>> cout<<arr->bval;
>>>> cout<<endl;
>>>> }
>>>> int main()
>>>> {
>>>> base BaseArr[5];
>>>> SomeFunc(BaseArr,5);
>>>> deri DeriArr[5];
>>>> SomeFunc(DeriArr,5);
>>>> system("pause");
>>>> return 0;
>>>> }
>>>>
>>>> 00000
>>>> 01010
>>>> Will somebody tell me why it's showing this..??
>>>>
>>>> --
>>>> With regards
>>>> ............
>>>> Puneet
>>>>
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>>>
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>>
>>
>>
>> --
>> With regards
>> ............
>> Puneet
>>
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>
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