You are right. He was really careful in picking double instead of int for this e...

ridiculous_fish · on Dec 23, 2014

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.