Hi your intention was logical OR or BITWISE OR ?

u did Logical.
Sanju
:)



On Sun, Aug 21, 2011 at 3:30 AM, sarvesh saran
<[email protected]>wrote:

> Hi Nitin,
>
> thanks that makes sense. I will try that out.
>
> I have another question. Is there a  really fast way of converting a
> hexadecimal string say "02F9A" to its decimal representation in C++?
>
> thanks,
> Sarvesh
>
> thanks,
> Sarvesh
>
>
> On Sun, Aug 21, 2011 at 3:41 PM, Nitin Nizhawan 
> <[email protected]>wrote:
>
>> int num = 0;
>> for(int i=0;i<A.size();i++){
>>    num=num||(A[i]<<3*i);
>> }
>> printf("%d",num);
>>
>> I think this will do. Given the number is with in the range of integer.
>>
>>
>> On Sun, Aug 21, 2011 at 3:40 PM, Nitin Nizhawan <[email protected]
>> > wrote:
>>
>>> int num = 0;
>>> for(int i=0;i<A.size();i++){
>>>    num=num||(A[i]<3*i);
>>> }
>>> printf("%d",num);
>>>
>>> I think this will do.
>>>
>>>
>>> On Sun, Aug 21, 2011 at 2:25 PM, sarvesh saran <
>>> [email protected]> wrote:
>>>
>>>> Hi,
>>>>
>>>> I have a vector<int> A or an array (for C guys) that contains the octal
>>>> representation of a number.
>>>>
>>>> So the array can be something like: [1,5,7] or [7,7,5,6,3,4,2] etc
>>>>
>>>> i.e no number in the array can be >= 8.
>>>>
>>>> Now given this array, I need to convert it its decimal representation.
>>>>
>>>> The naive way to do it would be to scan array from left to right, take
>>>> each digit, multiply by 8 pow (x) where x is from 0 to ...n and compute 
>>>> sum.
>>>>
>>>> i.e something like:
>>>>
>>>> int oct = 1;
>>>> int num = 0;
>>>>
>>>>  for(<array length>){
>>>>         num+= oct * A[i];
>>>>         oct = oct * 8;
>>>>     }
>>>>
>>>> is there a faster way to do this? maybe using some STL container or 
>>>> algorithm. ?
>>>>
>>>> thanks,
>>>> sarvesh
>>>>
>>>>
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>>>
>>>
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