From: nobu@...
Date: 2020-10-10T06:54:58+00:00
Subject: [ruby-core:100355] [Ruby master Bug#17257] Integer#pow(0, 1) returns 1, which is incorrect

Issue #17257 has been updated by nobu (Nobuyoshi Nakada).

Assignee set to mrkn (Kenta Murata)
Status changed from Open to Assigned

I'm not sure if it should be 0.
As `x.pow(y, m)` equals multiplying `x % m` to 1 `y` times, isn't 1 ok when `x == 0`?

----------------------------------------
Bug #17257: Integer#pow(0, 1) returns 1, which is incorrect
https://bugs.ruby-lang.org/issues/17257#change-87967

* Author: universato (Yoshimine Sato)
* Status: Assigned
* Priority: Normal
* Assignee: mrkn (Kenta Murata)
* Backport: 2.5: UNKNOWN, 2.6: UNKNOWN, 2.7: UNKNOWN
----------------------------------------
Ruby 2.5.8, 2.6.6, 2.7.1


```ruby
p -1.pow(0, 1) #=> 1
p  0.pow(0, 1) #=> 1
p  1.pow(0, 1) #=> 1
p 1234567890.pow(0, 1) #=> 1
```

These return values should be 0.


Patch for test:

Let's add some boundary value tests to `test_pow` of [test_numeric.rb](https://github.com/ruby/ruby/blob/e014e6bf6685f681998238ff005f6d161d43ce51/test/ruby/test_numeric.rb).

```ruby
integers = [-2, -1, 0, 1, 2, 3, 6, 1234567890123456789]
integers.each do |i|
  assert_equal(0, i.pow(0, 1), '[Bug #17257]')
  assert_equal(1, i.pow(0, 2))
  assert_equal(1, i.pow(0, 3))
  assert_equal(1, i.pow(0, 6))
  assert_equal(1, i.pow(0, 1234567890123456789))
  
  assert_equal(0,  i.pow(0, -1))
  assert_equal(-1, i.pow(0, -2))
  assert_equal(-2, i.pow(0, -3))
  assert_equal(-5, i.pow(0, -6))
  assert_equal(-1234567890123456788, i.pow(0, -1234567890123456789))
end

assert_equal(0,  0.pow(2, 1))
assert_equal(0,  0.pow(3, 1))
assert_equal(0,  2.pow(3, 1))
assert_equal(0, -2.pow(3, 1))



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