Well, first, whether swap() is used or not depends on the algorithm; second, the result of the sorting, as opposed to reversal, does not necessary look like the elements were swapped.
Oh, based on your other comment, I finally understood what you meant initially.
You were saying that the final array doesn't have many indices i,j such that a[i] = sorted_a[j] and a[j] = sorted_a[i].
My initial comment was not referring to the final order of the array, but the operations made to reach that order (one or multiple swaps). Another example could have been saying that we could have named the "sort" method "compare" because it uses comparisons in its algorithm (which was a parallel to your initial comment that it's called "swap" because the reverse operation uses swaps to achieve this).
> whether swap() is used or not depends on the algorithm
I didn't mean the std::swap() function itself, but the swapping of two elements a[i] and a[j]. Is there any sorting algorithm that doesn't rely on swapping two elements (or two parts of the array)? I guess, only if the sort is not being done in-place and the result is stored in a different variable/memory.
> does not necessary look like the elements were swapped
The only way elements don't look like they are swapped is elements are changed, added or removed, which doesn't happen during sort.
There are in-place algorithms that use shifting rather than swapping (the so called in-place merge sort, for instance).
If you look at the result and at the original, generally you will not find many pairs in which the elements exchanged their positions. It takes special initial arrangements for this to be true for all elements.
