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What is the output of this sort function?

Posted on 2008-06-10
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Last Modified: 2010-04-21
IF
vector<int>   A;   // 10 elements, say 0-9
vector<long>   B;  // 10   elements, say all == 1
AND
struct SPredicate
{
      const long* _rowBegin;
      SPredicate(const vector<long>& v) : _rowBegin( v.begin() ) { }
      bool operator()(int i, int j) { return _rowBegin[ i ]  <  _rowBegin[ j ] ; }
};
What would the ouput be of the following?
std::sort(A.begin(), A.end(),SPredicate(B));


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Question by:CTPAS
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Expert Comment

by:evilrix
ID: 21751489
The predicate is indexing B and comparing B using the current value of i and j (both of which will be values of A); however, all elements in B are 1 so I'd expect that A would remain in the same order as it was initially.
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Expert Comment

by:evilrix
ID: 21751551
^^^this assumes that the iterator of a vector is convertible to a type T *, which the C++ Standard doesn't as far as I can remember, define.
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Author Comment

by:CTPAS
ID: 21751616
Thanks for quick resonse
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Accepted Solution

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evilrix earned 125 total points
ID: 21751737
I believe this example code implements what your Q asks, yes? The output is what I stated I'd expect in my initial response, "...I'd expect that A would remain in the same order as it was initially"

Do you understand why this is the case though?
#include <vector>

#include <algorithm>

#include <iostream>
 

typedef std::vector<int> intvec_t;

typedef std::vector<long> longvec_t;
 

struct SPredicate

{

	longvec_t::const_iterator _rowBegin; 

	SPredicate(longvec_t & v) : _rowBegin( v.begin() ) { } 

	bool operator()(int i, int j) { return _rowBegin[ i ]  <  _rowBegin[ j ] ; } 

};
 
 

int main()

{

	int  aA[] = {5,6,7,8,9,0,1,2,3,4};

	long aB[] = {1,1,1,1,1,1,1,1,1,1};
 

	intvec_t A(aA, (aA+10));

	longvec_t B(aB, (aB+10));
 

	std::sort(A.begin(), A.end(), SPredicate(B));
 

	std::copy(A.begin(), A.end(), std::ostream_iterator<int>(std::cout, " "));

	std::cout << std::endl;
 

	std::copy(B.begin(), B.end(), std::ostream_iterator<long>(std::cout, " "));

	std::cout << std::endl;

}

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Author Closing Comment

by:CTPAS
ID: 31465747
yes. I was only unclear about where i , j came from in the comparison.
This method is a compact way to order set A elements based on the ordering of set B
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