You can also use .count, that returns 0 or 1, it is a sort of "is_there?"

2011/5/13 Leopoldo Taravilse <[email protected]>:
> I'm really sorry, my mistake. I always get a compile error while I write
> .find and I never remember the correct way of how it works.
>
> On Fri, May 13, 2011 at 2:20 PM, Axel Freyn <[email protected]> wrote:
>>
>> Just a small remark:
>>
>> On Fri, May 13, 2011 at 6:56 PM, Leopoldo Taravilse <[email protected]>
>> wrote:
>>>
>>> Here's an example
>>>
>>> set<int> setint; // you create a set
>>> setint.insert(4); // you add 4 to the set in o(log n) where n is the
>>> number of elements of the set
>>> setint.insert(5); // you add 5 to the set
>>> setint.insert(4); // nothing happens because 4 is already in the set
>>> if(setint.find(4)) cout << "4 is in the set" << endl; // you check in
>>> o(log n) if 4 is in the set, and because it is in the set you print "4 is in
>>> the set".
>>> setint.clear(); // you clear the set in o(1).
>>
>> setint.find(4)
>> returns an iterator -- so you can't use it in an if-clause. You should
>> write
>>  if(setint.find(4) != setfind.end() )
>> instead (if the key 4 is NOT found, "find" will compare equal to
>> setfind.end())
>>
>>
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-- 
Walter Erquínigo Pezo

Every problem has a simple, fast and wrong solution.

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