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C++ trying to swap values in a vector

This is my swap function:

template <typename t>
void swap (t& x, t& y)
{
    t te开发者_如何学Pythonmp = x;
    x = y;
    y = temp;
    return;
}

And this is my function (on a side note v stores strings) call to swap values but whenever I try to call using values in a vector I get an error. I'm not sure what I'm doing wrong.

swap(v[position], v[nextposition]); //creates errors


I think what you are looking for is iter_swap which you can find also in <algorithm>.
all you need to do is just pass two iterators each pointing at one of the elements you want to exchange.
since you have the position of the two elements, you can do something like this:

// assuming your vector is called v
iter_swap(v.begin() + position, v.begin() + next_position);
// position, next_position are the indices of the elements you want to swap


Both proposed possibilities (std::swap and std::iter_swap) work, they just have a slightly different syntax. Let's swap a vector's first and second element, v[0] and v[1].

We can swap based on the objects contents:

std::swap(v[0],v[1]);

Or swap based on the underlying iterator:

std::iter_swap(v.begin(),v.begin()+1);

Try it:

int main() {
  int arr[] = {1,2,3,4,5,6,7,8,9};
  std::vector<int> * v = new std::vector<int>(arr, arr + sizeof(arr) / sizeof(arr[0]));
  // put one of the above swap lines here
  // ..
  for (std::vector<int>::iterator i=v->begin(); i!=v->end(); i++)
    std::cout << *i << " ";
  std::cout << std::endl;
}

Both times you get the first two elements swapped:

2 1 3 4 5 6 7 8 9


There is a std::swap in <algorithm>


after passing the vector by reference

swap(vector[position],vector[otherPosition]);

will produce the expected result.


  1. Using std::swap by including the <algorithm> library to swap values by references / smart pointers,
    e.g. std::swap(v[0], v[1])
    Note: v[i] is a reference

  2. Using std::iter_swap from the same library,
    e.g. std::iter_swap(v.begin(), v.begin() + v.size() - 1)

  3. Using lvalue references and rvalue tuple by including the <tuple> library
    e.g. std::tie(v[0], v[1]) = std::make_tuple(v[1], v[0])
    Note: constexpr since C++14

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