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    <p>There is no difference between <code>this.method()</code> and <code>super.method()</code> until the said <code>method()</code> gets <em>overridden</em> in the caller's class.</p> <p>For example, with</p> <pre class="lang-java prettyprint-override"><code>class SuperClass { public void method() { System.out.println("SuperClass"); } } class SubClass extends SuperClass { public SubClass() { method(); this.method(); super.method(); } } </code></pre> <p>Calling</p> <pre class="lang-java prettyprint-override"><code>new SubClass(); </code></pre> <p>Prints</p> <pre class="lang-none prettyprint-override"><code>SuperClass SuperClass SuperClass </code></pre> <p>While with</p> <pre class="lang-java prettyprint-override"><code>class SuperClass { public void method() { System.out.println("SuperClass"); } } class SubClass extends SuperClass { @Override public void method() { System.out.println("SubClass"); } public SubClass() { method(); this.method(); super.method(); } } </code></pre> <p>Calling</p> <pre><code>new SubClass(); </code></pre> <p>Prints</p> <pre class="lang-none prettyprint-override"><code>SubClass SubClass SuperClass </code></pre> <hr> <p>In parallel, there is no difference between <code>this.field</code> and <code>super.field</code> until the said <code>field</code> gets <em>hidden</em> in the caller's class.</p> <p>For example, with</p> <pre class="lang-java prettyprint-override"><code>class SuperClass { protected String field = "SuperClass"; } class SubClass extends SuperClass { public SubClass(String field) { System.out.println(field); System.out.println(this.field); System.out.println(super.field); } } </code></pre> <p>Calling</p> <pre class="lang-java prettyprint-override"><code>new SubClass("parameter"); </code></pre> <p>Prints</p> <pre class="lang-none prettyprint-override"><code>parameter SuperClass SuperClass </code></pre> <p>While with</p> <pre class="lang-java prettyprint-override"><code>class SuperClass { protected String field = "SuperClass"; } class SubClass extends SuperClass { private String field = "SubClass"; public SubClass(String field) { System.out.println(field); System.out.println(this.field); System.out.println(super.field); } } </code></pre> <p>Calling</p> <pre><code>new SubClass("parameter"); </code></pre> <p>Prints</p> <pre class="lang-none prettyprint-override"><code>parameter SubClass SuperClass </code></pre> <hr> <p>Side note: <code>methods()</code> get <em>overriden</em> while <code>fields</code> get <em>hidden</em>.</p>
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    1. POWhy using "this" keyword to get superclass methods
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    1. CO+1 for the concise key statement "there is no difference between this.method() and super.method() until the said method get overridden in the caller's class."
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      1. USTheBlastOne
    2. COAll that is true, but can be shortened to: If you want to call the superclassmethod, use `super`, otherwise do nothing as by default (without using `this`) always the "most current implementation" will be used.
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      1. USThorsten Dittmar
    3. COThanks man. Finally I got it
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      1. USgjman2
 

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