I like what you're trying to do, but the wording used to introduce the
names `cr` and `cp` feels a bit contorted. How about using a single
name, c, introduced up front in this line:
+ * of the argument. Equality of a component {@code c} is determined
as follows:
with the rest of the text updated accordingly. Note the use of the
indefinite article `a component` when the name is introduced, and the
definite article `the component` in subsequent references.
-- Jon
On 1/6/20 1:08 PM, Joe Darcy wrote:
Hello,
Please review a clarification to the specification of Record.equals:
JDK-8234783: Improve wording of spec of Record.equals
http://cr.openjdk.java.net/~darcy/8234783.0/
Patch below.
Thanks,
-Joe
--- old/src/java.base/share/classes/java/lang/Record.java 2020-01-06
13:00:52.179585999 -0800
+++ new/src/java.base/share/classes/java/lang/Record.java 2020-01-06
13:00:51.987681999 -0800
@@ -105,9 +105,21 @@
* The implicitly provided implementation returns {@code true} if
and
* only if the argument is an instance of the same record type as
this object,
* and each component of this record is equal to the
corresponding component
- * of the argument, according to {@link
java.util.Objects#equals(Object,Object)}
- * for components whose types are reference types, and according
to the semantics
- * of the {@code equals} method on the corresponding primitive
wrapper type.
+ * of the argument. Equality of the components is determined as
follows:
+ * <ul>
+ *
+ * <li> If the component is of a reference type, the component
{@code
+ * cr} is considered equal if and only if {@link
+ * java.util.Objects#equals(Object,Object)
+ * Objects.equals(this.cr(), r.cr()} would return true .
+ *
+ * <li> If the component is of a primitive type, using the
+ * corresponding primitive wrapper class {@code P} the component
+ * {@code cp} is considered equal if and only if {@code
+ * P.valueOf(this.cp()).equals(P.valueOf(r.cp()))} would return
+ * true.
+ *
+ * </ul>
*
* @see java.util.Objects#equals(Object,Object)
*
