StackOverflow2013

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Id
12653062
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2012-09-29T13:24:51.017
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2012-09-29T14:09:00.070
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2012-09-29T14:09:00.070
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1679863
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12652899
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2
Score
8
ViewCount
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LastEditorDisplayName
text
Body
<ul> <li>Packages provide logical namespace to your classes..</li> <li>And these packages are stored in the form of directory levels (they are converted to nested directories) to provide physical grouping (namespace) to your classes..</li> </ul> Also, note that the <code>physical</code> namespace has to be in accordance with the <code>logical</code> namespace.. You can't have your class with <code>package com.demo</code>, under directory structure : - <code>\com\demo\temp\</code>, it has to be under <code>\com\demo\</code>, and this directory is added to the classpath so that your classes is visible to JVM when it runs your code.. Suppose you have following directory structure: - <blockquote> A | +-- Sample.java(contains Demo class under package B) | +-- Outer.java(contains Demo class - no package) | +--B |    | |   +-- Demo.class | +--C |    | |   +-- Abc.class | +-- Demo.class </blockquote> Suppose, your class <code>Abc.class</code> and <code>Demo.class</code> (under directory A), isn't define under any package, whereas your class <code>Demo.class</code> (under directory B) is defined under <code>package B</code>. So, you need to have two directories in your classpath: - <code>\A</code>(for two classes: - <code>Demo.class</code> and <code>B.Demo.class</code>) and <code>\A\C</code>(for class <code>Abc.class</code>).. <ul> <li>Classes under different package can use same name. That is why there won't be any conflict between the two <code>Demo.class</code> defined above.. Because they are in different <code>packages</code>. That is the whole point of dividing them into <code>namespaces</code>.. This is beneficial, because you will not run out of unique names for your classes..</li> </ul>
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1. PODifference between Package and Directory in Java
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1. PTAnswer
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1. USRohit Jain
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1. USRohit Jain
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