Subtracting two vector<point2f> directly

Question

I have to subtract two vectors of type Point2f (both are of same size). I know it can be done by extracting the values at each index and then subtracting them in a loop but is there a direct method to it? Something like

Vector<Point2f> A[3];
A[2] = A[1] - A[0];

Answer 1

just for sports ;)

std::vector<Point2f> A,B;
A.push_back(Point2f(1,2));
A.push_back(Point2f(3,4));
A.push_back(Point2f(5,6));
B.push_back(Point2f(5,2));
B.push_back(Point2f(4,4));
B.push_back(Point2f(3,6));

// Mat C; subtract(A,B,C);
Mat C = Mat(A) - Mat(B);
cout<< A << endl << B << endl <<C<<endl;



[1, 2;  3, 4;  5, 6]
[5, 2;  4, 4;  3, 6]
[-4, 0;  -1, 0;  2, 0]

Answer 2

As per the documentation link that you provided, subtraction of two points is supported. So the following should work:

std::transform (A[1].begin(), A[1].end(), A[0].begin(), A[2].begin(), std::minus<Point2f>());

Note that this assumes that A[2] is big enough to store the result.

Alternative, you could write your own overloaded operator-() for vector subtraction:

const vector<Point2f> operator-(const vector<Point2f>& lhs, const vector<Point2f>& rhs)
{ ... }

You would need to optimize the above function to avoid a copy of the vector when the function returns. This does not preclude the need to write the looping code that you want to avoid. But it will give you a cleaner syntax for vector subtractions.