C++; Floating Point exception; Without ./ operator

Question

The example below, is from OpenCv documentation[1].

Mat H(100, 100, CV_64F);
for(int i = 0; i < H.rows; i++)
    for(int j = 0; j < H.cols; j++)
        H.at<double>(i,j)=1./(i+j);

This works perfectly fine. But in the last line what is ./ operator? And if I replace it with / it gives me floating point exception.

So, in both cases we have infinity when i and j are 0; then why do we get floating point exception for the second case?

[1] http://docs.opencv.org/trunk/d3/d63/classcv_1_1Mat.html

If you got it, I could delete my answer so you'll be able to delete you post. Just say the word. — StoryTeller - Unslander Monica, Sep 30 '17 at 20:14
@StoryTeller I think its a good answer, please don't delete it. It would help the community. I am waiting for few more answers. Why do you want to delete it (i am just curious)? — Porcupine, Sep 30 '17 at 20:17
Had you wanted to delete your question, my post would prevent it because it's upvoted. So just wanted to give you the option if this attracts a heap of downvotes. — StoryTeller - Unslander Monica, Sep 30 '17 at 20:20
I can't reproduce `floating-point-exception` for code as supplied -- even if changing `./` to `/`. Nevertheless, I suggest you replace the assignment with a `cerr<<` in order to see what values are generated in each case. — Brent Bradburn, Sep 30 '17 at 20:43
...I'm guessing the error was generated when you excluded the `+1` in `(i+j+1)`. — Brent Bradburn, Sep 30 '17 at 20:46
@nobar Yes you are right it was generated when i did 1/(i+j) — Porcupine, Sep 30 '17 at 20:47
@nobar So, in both cases we have infinity when i and j are 0; then why do we get floating point exception for the second case? — Porcupine, Sep 30 '17 at 20:50
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/155675/discussion-between-nikhil-and-nobar). — Porcupine, Sep 30 '17 at 20:57

score 4 · Answer 1 · answered Sep 30 '17 at 20:07

4

./ is not an operator. The dot binds with the 1, making it a double constant. It's equivalent to this:

1.0 / (i+j+1)

Only a bit shorter.

When you omit the dot, the expression is evaluated using integer arithmetic, giving all zero for all entries but 0, 0.

answered Sep 30 '17 at 20:07

StoryTeller - Unslander Monica

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Less well known is the /. operator, which open sources your website. – Yakk - Adam Nevraumont Oct 01 '17 at 00:19

Ron · Answer 2 · 2017-09-30T20:34:43.330

The . character is part of the 1. double literal. / is an arithmetic operator so the right hand side expression becomes:

1. / (i+j+1);

and the result is a value of type double. Omitting the . character makes it an integer literal of 1 and the expression becomes:

1 / (i+j+1);

where both operands are integer values and the result is an integer value. Spaces in C++ code make no difference to compiler. For readability reasons the statement should include spaces where appropriate:

H.at<double>(i,j) = 1. / (i+j+1);

C++; Floating Point exception; Without ./ operator

2 Answers2