Question

From the OpenCV documentation, it appears that copying a matrix is done using a shallow copy, but when changing one of the copies, a copy is done.

The exact reference is:

Mat& Mat::operator = (const Mat& m)

Mat& Mat::operator = (const MatExpr_Base& expr)

Mat& operator = (const Scalar& s)

Matrix assignment operators

Parameters:

m – The assigned, right-hand-side
matrix. Matrix assignment is O(1)
operation, that is, no data is copied.
Instead, the data is shared and the
reference counter, if any, is
incremented. Before assigning new
data, the old data is dereferenced via
Mat::release .

expr – The assigned matrix expression
object. As opposite to the first form
of assignment operation, the second
form can reuse already allocated
matrix if it has the right size and
type to fit the matrix expression
result. It is automatically handled by
the real function that the matrix
expressions is expanded to. For
example, C=A+B is expanded to
cv::add(A, B, C) , and add() will take
care of automatic C reallocation.

s – The scalar, assigned to each
matrix element. The matrix size or
type is not changed.

However, this appears not to work

Mat_<float> a(5,5),b(5,5);
a =1;
b = a;
a = 2;

now b == 2, intead of 1

Answer 1

It seems you misunderstood. “Before assigning new data, the old data is dereferenced via Mat::release” does not mean that when you write on a or b then a copy occurs. It means that when you type b=a, you lose the data that was in b.

Long story short : copy on write is not supported.