[ruby-core:99194] [Ruby master Bug#17017] Range#max & Range#minmax incorrectly use Float end as max

[merch-redmine@jeremyevans.net]

Issue #17017 has been updated by jeremyevans0 (Jeremy Evans).


I've added a pull request to restore compatibility for Range#max with integer beginning and Float::Infinity end: https://github.com/ruby/ruby/pull/3326

I think this pull request should require matz approval because it is a deliberate tradeoff of correctness in order to keep compatibility.  The correct behavior should be raising RangeError, as that is what an endless range returns.  If you would like matz to consider it, please bring it up for discussion at a future Ruby developers meeting.

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Bug #17017: Range#max & Range#minmax incorrectly use Float end as max
https://bugs.ruby-lang.org/issues/17017#change-86574

* Author: sambostock (Sam Bostock)
* Status: Open
* Priority: Normal
* ruby -v: ruby 2.8.0dev (2020-07-14T04:19:55Z master e60cd14d85) [x86_64-darwin17]
* Backport: 2.5: UNKNOWN, 2.6: UNKNOWN, 2.7: UNKNOWN
----------------------------------------
While continuing to add edge cases to [`Range#minmax` specs](https://github.com/ruby/spec/pull/777), I discovered the following bug:

```ruby
(1..3.1).to_a        == [1, 2, 3] # As expected

(1..3.1).to_a.max    == 3         # As expected
(1..3.1).to_a.minmax == [1, 3]    # As expected

(1..3.1).max    == 3.1            # Should be 3, as above
(1..3.1).minmax == [1, 3.1]       # Should be [1, 3], as above
```

One way to detect this scenario might be to do (whatever the C equivalent is of)

```ruby
range_end.is_a?(Numeric)                      // Is this a numeric range?
  && (range_end - range_begin).modulo(1) == 0 // Can we reach the range_end using the standard step size (1)
```

As for how to handle it, a couple options come to mind:

- We could error out and do something similar to what we do for exclusive ranges

```ruby
raise TypeError, 'cannot exclude non Integer end value'
```

- We might be able to calculate the range end by doing something like

```ruby
num_steps = (range_end / range_beg).to_i - 1 # one fewer steps than would exceed the range_end
max = range_beg + num_steps                  # take that many steps all at once
```

- We could delegate to `super` and enumerate the range to find the max

```ruby
super
```

- We could update the documentation to define the max for this case as the `range_end`, similarly to how the documentation for `include?` says it behaves like `cover?` for numeric ranges.



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