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Comment by pavanky | Hacker News Reader

The C greater function is unnecessarily long

  int greater(const void* p, const void* q)
  {
     return *(double *)p - *(double *)q;
  }

would work just as well.

EDIT: DON'T USE THIS, WONT ALWAYS WORK

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That is technically undefined behavior, because the result of the double subtraction could very well be a number that is not representable within the range of a signed int.

What is the point of defining the function with two void* arguments if the cast will just take place anyway?

The signature of greater is relevant because it has to conform to the interface for qsort. See the final argument of the interface here: http://www.cplusplus.com/reference/cstdlib/qsort/

Don't do this. It has overflow issues.

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

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.

">" 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.

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