What does this statement 'p = I.ptr(i);' do?

Question

Apparently, the search engine doesn't seem to be taking this symbol ' <> ' into search, Hence can anyone please explain to me what does this symbol mean as well as the statement?

For the statement, I can guess that

p = I.ptr<uchar>(i);

p happens to be pointing at I[i] address.

Thanks :)

Please [read the doc](http://docs.opencv.org/master/d3/d63/classcv_1_1Mat.html#a5a9ffc908ac90604f36a8b6a1038747d) before posting such questions. BTW, `p` points to the beginning of the `i`-th row of `I`, not to `I[i]` — Miki, Jul 04 '17 at 18:48

Curious · Accepted Answer · 2017-07-04T18:30:08.377

0

It's calling a member function of the class with an argument of i. For example

#include <iostream>

using std::cout;
using std::endl;

class Something {
public:
    template <typename T>
    T do_something(T in) {
        cout << __PRETTY_FUNCTION__ << endl;
        return in;
    }
};

int main() {
    auto something = Something{};
    // <int> not really needed here, the type can be deduced,
    // just here for demonstration purposes
    auto integer = something.do_something<int>(1);
    (void) integer;
}

Note that in order to do this in the context of a template type you would need to prefix the method call with template as explained in this question - Where and why do I have to put the "template" and "typename" keywords?

edited Jul 04 '17 at 18:30

answered Jul 04 '17 at 18:27

Curious

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so ptr is a function of class I and argument passed is i? What does that do in your code? – MaNyYaCk Jul 04 '17 at 18:29
@MaNyYaCk `ptr` is a method of the class, and the argument passed is `i` yes. the `int` in my code does the same thing as `i` in yours. Although the `` is not needed, the type can be deduced – Curious Jul 04 '17 at 18:31
so the in my code is basically a typecasting? – MaNyYaCk Jul 04 '17 at 18:36
@MaNyYaCk it's supplying a type parameter to a template method – Curious Jul 04 '17 at 18:41

What does this statement 'p = I.ptr(i);' do?

1 Answers1