How can matrix assignment in boost library work with parenthesis?

Question

From this sample code from Boost official site:

#include <boost/numeric/ublas/matrix.hpp>
#include <boost/numeric/ublas/io.hpp>

int main () {
    using namespace boost::numeric::ublas;
    matrix<double> m (3, 3);
    for (unsigned i = 0; i < m.size1 (); ++ i)
        for (unsigned j = 0; j < m.size2 (); ++ j)
            m (i, j) = 3 * i + j;
    std::cout << m << std::endl;
}

I am confused with m (i, j) = 3 * i + j; because m is a object and the only case that combines class and argument together is constructor function, but there, it is obviously not.

I am a beginner of C++. However, being different from Ruby there is few trick in C++.

In order to have a deep discovery toward C++, is there any who can give me a explanation in principle about that?

In c++ you can define your own operators, one of the most popular being `operator[]`, but `operator()` is also possible. Check this: https://www.boost.org/doc/libs/1_57_0/boost/numeric/ublas/matrix.hpp and search for `const_reference operator () (size_type i, size_type j)` — Frederik.L, Apr 08 '18 at 03:18
Also, all lvalues can be assigned. | You need to learn more about C++ before asking questions to [so] otherwise such questions would be considered unclear. — user202729, Apr 08 '18 at 03:18
@Frederik.L, thanks a lot. In my imagination, () is not a operator, because function call and new object. Now I have new understanding to C++. — pah8J, Apr 08 '18 at 05:06
@MartinS.Victory You are welcome. Since it seems to answer the question, I will post it as an answer. — Frederik.L, Apr 08 '18 at 05:33
You need to learn more about C++ [operator overloading](http://en.cppreference.com/w/cpp/language/operators) and probably [object](http://en.cppreference.com/w/cpp/language/object). — user202729, Apr 08 '18 at 05:45

Frederik.L · Accepted Answer · 2018-04-08T05:55:32.553

In C++, you can define your own operators (and override them if you need to). One popular operator for accessors is []. However, () is possible as well for a custom operator.

If you take look at the source code of matrix.hpp from Boost, in which the matrix object is defined, there is indeed an operator ().

/** Access a matrix element. Here we return a const reference
 * \param i the first coordinate of the element. By default it's the row
 * \param j the second coordinate of the element. By default it's the column
 * \return a const reference to the element
 */
    BOOST_UBLAS_INLINE
    const_reference operator () (size_type i, size_type j) const {
        return data () [layout_type::element (i, size1_, j, size2_)];
    }

and

/** Access a matrix element. Here we return a reference
 * \param i the first coordinate of the element. By default it's the row
 * \param j the second coordinate of the element. By default it's the column
 * \return a reference to the element
 */
    BOOST_UBLAS_INLINE
    reference operator () (size_type i, size_type j) {
        return at_element (i, j);
    }

    // Element assignment

The lower level implementation of Boost mechanics might be a little complex to understand at the first look, but what allows it to have a syntax like this is the presence of operator () in the definition.

You can check simpler examples about operators, for example there (on cppreference).

How can matrix assignment in boost library work with parenthesis?

1 Answers1