Modified min and max for code uniformity

[m16vie]

No description provided.

[gribozavr]

Does this change implement what was discussed in pull request #137?

[colinta]

This changes the behavior of max. If a class implements Comparable, calling max(a, b, c) currently returns the last instance that has the maximum value. This means that if a > b, b == c, the current version of max returns c, but your version returns b.

Neither is correct or wrong, but the change is unnecessary.

[m16vie]

Also, currently max(a, b, c) returns b assuming a < b, b == c and max(a, b, c, d) returns d assuming a < b, b == c == d, which seems to be wrong. Following the argument that given equal parameters, min and max should return different instances, max should then return the last maximum instance while min returns the first minimum. I suggest the following code.

/// Returns the greatest argument passed.
@warn_unused_result
public func max<T : Comparable>(x: T, _ y: T, _ z: T, _ rest: T...) -> T {
  var r = x
  if y >= r {
    r = y
  }
  if z >= r {
    r = z
  }
  for t in rest {
    if t >= r {
      r = t
    }
  }
  return r
}

[gribozavr]

@mmkg Please update the pull request and provide tests.

[IngCr3at1on]

@colinta I think the general idea was to create more consistency in the language handling in the first place (both are "valid" but currently they are inconsistent to my understanding). This was discussed in great detail on #137

[gribozavr]

#566 adds tests that show that we already implement #137.