confusion about using std::less and std::greater with std::sort

std::sort sorts in ascending order by default. In case you are looking for descending order, here's the trick:

int x[10] = { 1,3,5,7,9,2,4,6,8,10 };
std::vector<int> vec(x, x+10);          // construct std::vector object
std::sort(vec.rbegin(),vec.rend());     // sort it in reverse manner

This way, you explicitly say that std::sort should treat your array as its end is its beginning and vice versa, which results in your array being sorted in descending order. Here's the full example.

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And in case you want to use std::less and std::greater, then it could look like this:

int x[10] = { 1,3,5,7,9,2,4,6,8,10 };
std::sort(x, x + 10, std::less<int>());     // for ascending order
std::sort(x, x + 10, std::greater<int>());  // for descending order

Full example with second solution is here.

std::sort behaves like that because it's based on the idea of a strict weak ordering, which is (usually) defined in terms of the < operator.

As to your question; it currently seems to be "I wrote a C function that behaves differently to std::sort. Why is it different?". The answer is: because you wrote a different function!