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Re: parallel equal not parallel


On 11/04/2010 10:53 PM, Paolo Carlini wrote:
On 11/04/2010 09:48 PM, François Dumont wrote:
So it uses the normal mismatch algo, IMO it should be:

   template<typename _IIter1, typename _IIter2>
     inline bool
     equal(_IIter1 __begin1, _IIter1 __end1, _IIter2 __begin2)
     {
       return mismatch(__begin1, __end1, __begin2).first == __end1;
     }

   template<typename _IIter1, typename _IIter2, typename _Predicate>
     inline bool
     equal(_IIter1 __begin1, _IIter1 __end1, _IIter2 __begin2,
           _Predicate __pred)
     {
       return mismatch(__begin1, __end1, __begin2, __pred).first ==
__end1;
     }

Agree ?
Johannes?

Looks reasonable. I'm going to regression test this patch.


Johannes


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