CODEWITHSUNDEEP
c++sortingvectorstlrange
PapaDiHatti
PapaDiHatti

Reputation: 1921

Sorting part of vector (in place)

I could not understand why below piece of code is not sorting first two elements of vector :

int main() {
    std::vector<int> v = {2,1,3,1,2};
    std::sort(v.begin(),v.begin()+1);
    for(auto elem:v)
    {
      std::cout<<elem<<std::endl;
    }
    // your code goes here
    return 0;
}

Any thoughts ?

Upvotes: 2

Views: 3947

Answers (4)

Mansuro
Mansuro

Reputation: 4617

Sorts the elements in the range [first, last)

from std::sort

In case you are not familiar with the notation, ) means last is not included, you can find more information about that in this question.

Upvotes: 1

Eduard Rostomyan
Eduard Rostomyan

Reputation: 6546

In order to sort first n elements you must indicate you call as follows:

sort(V.begin(), V.begin() + n);

Thus for 2 elements you must call:

 sort(V.begin(), V.begin() + 2);

This is because all STL algorithms takes open range [first, last).

Upvotes: 1

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

This range v.begin(),v.begin()+1 that can be mathematically written like [v.begin(), v.begin() + 1 ) contains only one element v[0] that is equal to 2. If you want to sort a range of 2 elements then you should write

std::sort( v.begin(), std::next( v.begin(), 2 ) );

that is equivalent to

std;:sort( v.begin(), v.begin() + 2 );

Upvotes: 3

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170084

std::sort (and all standard library algorithms) expects a half-open range. The end iterator is a one-past end indicator (the open part). So [it, it + 1) is a range of just one element. In your case, it's just the first vector element.

And well, a one element range is already sorted.

Upvotes: 7

Related Questions