From: universato@... Date: 2020-10-10T09:18:59+00:00 Subject: [ruby-core:100356] [Ruby master Bug#17257] Integer#pow(0, 1) returns 1, which is incorrect Issue #17257 has been updated by universato (Yoshimine Sato). According to RDoC, `x.pow(y, m)` equals `(x**y) % m`. In fact, `x.pow(y, m)` returns same value as `(x**y) % m` except for `x.pow(0, 1)`. ```ruby p (12**0) % 1 #=> 0 p 12.pow(0, 1) #=> 1 ``` ---------------------------------------- Bug #17257: Integer#pow(0, 1) returns 1, which is incorrect https://bugs.ruby-lang.org/issues/17257#change-87968 * 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)) -- https://bugs.ruby-lang.org/ Unsubscribe: <mailto:ruby-core-request@ruby-lang.org?subject=unsubscribe> <http://lists.ruby-lang.org/cgi-bin/mailman/options/ruby-core>