Puzzle (Oddity)

Consider the following example.

public static boolean isOdd( int i ) {
  return i % 2 == 1;
}

Do you think that the above implementation is correct?

When in doubt, write a test.

package demo;

import org.junit.jupiter.params.ParameterizedTest;
import org.junit.jupiter.params.provider.ValueSource;

import static demo.App.isOdd;
import static org.junit.jupiter.api.Assertions.assertTrue;

public class AppTest {

  @ParameterizedTest( name = "should return true when given the odd value {0}" )
  @ValueSource( ints = { 1, -1 } )
  void shouldReturnTrueWhenGivenOdd( int oddNumber ) {
    assertTrue( isOdd( oddNumber ) );
  }
}

The above test will fail.

$ ./gradlew test

> Task :test FAILED

AppTest > should return true when given the odd value 1 PASSED

AppTest > should return true when given the odd value -1 FAILED
    org.opentest4j.AssertionFailedError at AppTest.java:14

This example was taken from PUZZLE 1: ODDITY in Java™ Puzzlers: Traps, Pitfalls, and Corner Cases.

“If you divide a by b, multiply the result by b, and add the remainder, you are back where you started JLS 15.17.3. This identity makes perfect sense, but in combination with Java’s truncating integer division operator JLS 15.17.2, it implies that when the remainder operation returns a nonzero result, it has the same sign as its left operand.”