Began testing class subtyping with mixins
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513bd1a733
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@ -186,4 +186,42 @@ require( 'common' ).testCase(
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"Crap; order matters?!"
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);
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},
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/**
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* If some trait T used by abstract class C defines abstract method M,
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* then some subtype C' of C should be able to provide a concrete
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* definition of M such that T.M() invokes C'.M.
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*/
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'Abstract method inherited from trait can be implemented by subtype':
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function()
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{
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var T = this.Sut(
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{
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'public doFoo': function()
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{
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// should invoke the concrete implementation
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this.foo();
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},
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'abstract protected foo': [],
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} );
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var called = false;
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// C is a concrete class that extends an abstract class that uses
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// trait T
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var C = this.AbstractClass.use( T ).extend( {} )
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.extend(
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{
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// concrete definition that should be invoked by T.doFoo
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'protected foo': function()
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{
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called = true;
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},
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} );
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C().doFoo();
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this.assertOk( called );
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},
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} );
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@ -0,0 +1,59 @@
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/**
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* Tests extending a class that mixes in traits
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*
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* Copyright (C) 2014 Mike Gerwitz
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*
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* This file is part of GNU ease.js.
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*
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* ease.js is free software: you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation, either version 3 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program. If not, see <http://www.gnu.org/licenses/>.
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*/
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require( 'common' ).testCase(
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{
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caseSetUp: function()
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{
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this.Sut = this.require( 'Trait' );
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this.Class = this.require( 'class' );
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},
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/**
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* The supertype should continue to work as it would without the
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* subtype, which means that the supertype's traits should still be
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* available. Note that ease.js does not (at least at the time of
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* writing this test) check to see if a trait is no longer accessible
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* due to overrides, and so a supertype's traits will always be
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* instantiated.
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*/
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'Subtype instantiates traits of supertype': function()
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{
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var called = false;
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var T = this.Sut(
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{
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foo: function() { called = true; },
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} );
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// C is a subtype of a class that mixes in T
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var C = this.Class.use( T ).extend( {} )
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.extend(
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{
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// ensure that there is no ctor-dependent trait stuff
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__construct: function() {},
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} );
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C().foo();
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this.assertOk( called );
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},
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} );
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@ -134,4 +134,53 @@ require( 'common' ).testCase(
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C().doFoo();
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this.assertOk( called );
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},
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/**
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* If a supertype mixes in a trait that provides a virtual method, a
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* subtype should be able to provide its own concrete implementation.
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* This is especially important to test in the case where a trait
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* invokes its own virtual method---we must ensure that the message is
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* properly passed to the subtype's override.
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*
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* For a more formal description of a similar matter, see the
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* AbstractTest case; indeed, we're trying to mimic the same behavior
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* that we'd expect with abstract methods.
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*/
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'Subtype can override virtual method of trait mixed into supertype':
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function()
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{
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var _self = this;
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var T = this.Sut(
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{
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'public doFoo': function()
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{
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// this call should be passed to any overrides
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return this.foo();
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},
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// this is the one we'll try to override
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'virtual protected foo': function()
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{
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_self.fail( true, false, "Method not overridden." );
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},
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} );
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var called = false;
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// C is a subtype of a class that implements T
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var C = this.Class.use( T ).extend( {} )
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.extend(
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{
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// this should be called instead of T.foo
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'override protected foo': function()
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{
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called = true;
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},
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} );
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C().doFoo();
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this.assertOk( called );
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},
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} );
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