Why does using += on a nullable type result in a FORWARD_NULL defect

Question

No doubt there are other, perhaps better ways to do this, but I'm trying to understand what is going on here.

In the below example, coverity is reporting a FORWARD_NULL defect on the fourth line.

double? foo = null;
double bar = 1.23;
foo += bar;
System.Windows.Point point = new System.Windows.Point(foo,bar);

it reports:

assign_zero: Assigning: foo = null.

on the foo += bar line.

in += Operator (C# Reference), I see that x += y is equivalent to x = x + y, and in Using nullable types (C+ Programming Guide), I see that

These operators [the binary operator] produce a null value if one or both operands are null

so is that what is going on? foo += bar becomes foo = foo + bar and since foo is null, foo + bar is null?

Is the *defect* being reported on line three, or is line three *evidence* for a defect that comes later? Normally the forward null defect is reported at the location where a null dereference can throw, but there's no such dereference here. — Eric Lippert, Mar 08 '19 at 16:34
@Eric Lippert you are correct. The defect is reported a few lines later where foo is being dereferenced. I'll update the question — Michael J., Mar 08 '19 at 20:02
OK good, but the program fragment you gave shouldn't even *compile*. `Point` takes a double, not a nullable double. There's no reason to run code through Coverity if it doesn't even compile! — Eric Lippert, Mar 08 '19 at 22:54

score 5 · Answer 1 · answered Mar 08 '19 at 15:16

5

so is that what is going on? foo += bar becomes foo = foo + bar and since foo is null, foo + bar is null?

Yes.

answered Mar 08 '19 at 15:16

spodger

1 Answers1