long and int equality, what goes on with Equals()

Question

I encountered this today, I am sure there is an excellent reason but why is is the case for num2 equal to num1 inconsistent?

long num1 = 1;
int num2 = 1;

num1.Equals(num2);  // TRUE - cool.
num2.Equals(num1); // FALSE - hmm?
num2 == num1; // TRUE - ?! Whit?

Why is this.

http://www.codeproject.com/Articles/584128/What-is-the-difference-between-equalsequals-and — Daniel Netzer, Jun 21 '16 at 18:32
@Daniel: Not quite. This question here is about confusion of implicit widening conversions. — Joey, Jun 21 '16 at 18:36

Lee · Answer 1 · 2016-06-21T18:39:38.567

14

num1.Equals(num2);

resolves to the overload of long.Equals(long) since there is an implicit widening conversion from int to long.

num2.Equals(num1);

calls int.Equals(object) which causes the long num1 to be boxed. Since this is not an int, the comparison returns false.

num2 == num1;

calls the overloaded == operator for long which again causes num2 to be promoted to a long before being compared.

edited Jun 21 '16 at 18:39

answered Jun 21 '16 at 18:35

Lee

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Yes absolutely, for value type it's not only sufficient to have similar value but types also must be same (or) should be implicit casted. – Rahul Jun 21 '16 at 18:37
@Rahul It has nothing to do with whether it's a value or reference type. Pretty much all (good) implementations of equality are defined to only return true for objects of the same type. To not do so is typically a bug in the implementation of `Equals`. – Servy Jun 21 '16 at 18:42

score 0 · Answer 2 · answered Jun 21 '16 at 18:37

long num1 = 1;
int num2 = 1;

MessageBox.Show(num1.Equals(num2).ToString()); 
MessageBox.Show(num2.Equals((int)num1).ToString());
MessageBox.Show((num2 == num1).ToString());

It's a matter of explicitly casting the long as an int to make them equal as expected. The other ones are doing the casting for you.

long and int equality, what goes on with Equals()

2 Answers2