Actually I would say it was rather careless, because it mishandles NaNs: vector<... | Hacker News

Hacker Newsnew | past | comments | ask | show | jobs | submit

ridiculous_fish on Dec 23, 2014 | parent | context | favorite | on: Popular Myths about C++, Part 3

Actually I would say it was rather careless, because it mishandles NaNs:

    vector<double> v = {0, nan(""), 1};
    sort(v.begin(), v.end(), [](double x, double y) { return x>y; });

In my test this leaves the vector unchanged, which is definitely not in decreasing order.

lomnakkus on Dec 23, 2014 [–]

">" does not form a total order on doubles if you include NaN, so that's not really suprising. That is, you're providing a comparison function which is invalid, you should (at best) expect invalid results.

Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact