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JShell error "unexpected type" when using specific class name

Tags:

java

jshell

I was just playing around with JShell, and it seems that defining class Z{} and then defining var z = new Z() does not work. But using different class names, like class X and class A, does work.

Surely I must be missing something obvious...?

|  Welcome to JShell -- Version 14.0.1
|  For an introduction type: /help intro

jshell> class X{}
|  created class X

jshell> class Z{}
|  created class Z

jshell> var x = new X()
x ==> X@26a1ab54
|  created variable x : X

jshell> var z = new Z()
|  Error:
|  unexpected type
|    required: class
|    found:    type parameter Z
|  var z = new Z();
|              ^

jshell> class A{}
|  created class A

jshell> var a = new A()
a ==> A@2ef1e4fa
|  created variable a : A
like image 430
user140547 Avatar asked May 30 '20 12:05

user140547


1 Answers

Use of var can lead to a variable having a non-denotable type. For example, looking at the return type of an expression that could be String or Integer:

jshell> /set feedback verbose
jshell> var x = true ? "a" : 1
x ==> "a"
|  created variable x : Serializable&Comparable<? extends Serializable&Comparable<?>&java.lang.constant.Constable&java.lang.constant.ConstantDesc>&java.lang.constant.Constable&java.lang.constant.ConstantDesc

When jshell is evaluating your code fragment, if this is the case, it wraps it in a block of code so that it can record this type for later use. The wrapping fragment includes a generic type parameter called Z:

        // private static <Z> Z do_itAux() {
        //     wtype x_ = y;
        //     @SuppressWarnings("unchecked")
        //     Z x__ = (Z) x_;
        //     return x__;

This parameter's name leaks into the code block being evaluated, meaning the class's name is shadowed by the type parameter. This makes Z a special case -- the other single character examples are fine.

like image 175
Joe Avatar answered Nov 12 '22 22:11

Joe