Math operations with null

Question

Please explain why this test passes?

[Test] public void TestNullOps() {     Assert.That(10 / null, Is.Null);     Assert.That(10 * null, Is.Null);     Assert.That(10 + null, Is.Null);     Assert.That(10 - null, Is.Null);     Assert.That(10 % null, Is.Null);     Assert.That(null / 10, Is.Null);     Assert.That(null * 10, Is.Null);     Assert.That(null + 10, Is.Null);     Assert.That(null - 10, Is.Null);     Assert.That(null % 10, Is.Null);      int zero = 0;     Assert.That(null / zero, Is.Null); } 

I don't understand how this code even compiles.

Looks like each math expression with null returns Nullable<T> (e.g. 10 / null is a Nullable<int>). But I don't see operator methods in Nullable<T> class. If these operators are taken from int, why the last assertion doesn't fail?

Answer 1

From MSDN:

The predefined unary and binary operators and any user-defined operators that exist for value types may also be used by nullable types. These operators produce a null value if the operands are null; otherwise, the operator uses the contained value to calculate the result.

That's why all the test are passed, including the last one - no matter what the operand value is, if another operand is null, then the result is null.

Answer 2

The operators for Nullable<T> are so-called "lifted" operators]; the c# compiler takes the operators available for T and applies a set of pre-defined rules; for example, with +, the lifted + is null if either operand is null, else the sum of the inner values. Re the last; again, division is defined as null if either operand is null - it never performs the division.