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    <p><strong>Why are you doing that?</strong> <em>(exec apart)</em><br> <strong>Why are you trying to do that with <code>exec</code>?</strong></p> <p>Also, doing it with <code>exec</code> will:</p> <ol> <li>Not work.</li> <li>Give different results in <code>python-2.x</code> and in <code>python-3.x</code>.</li> </ol> <p>Example:</p> <pre><code>class demo: a = 'a' print(demo.a) def start(): global demo class demo: b = "b" try: print(demo.a) except AttributeError: print(demo.b) if __name__ == '__main__': start() try: print(demo.a) except AttributeError: print(demo.b) </code></pre> <p>That either in <code>python-2.x</code> and in <code>python-3.x</code> will give:</p> <pre><code>a b b </code></pre> <p>And now let's try it with <code>exec</code>:</p> <pre><code>class demo: a = 'a' print(demo.a) def start(): exec('global demo', globals(), locals()) exec('class demo: b = "b"', globals(), locals()) try: print(demo.a) except AttributeError: print(demo.b) if __name__ == '__main__': start() try: print(demo.a) except AttributeError: print(demo.b) </code></pre> <p>Output <code>python2.7</code>:</p> <pre><code>a b a </code></pre> <p>Output <code>python3.2</code>:</p> <pre><code>a a a </code></pre> <hr> <p><strong>Q: How to 'dynamically create the class'?</strong></p> <p>As <em>kindall</em> already told you, <code>exec</code> is not the way to do that.</p> <p>A <a href="https://stackoverflow.com/q/100003/1132524">metaclass</a> or a <a href="https://stackoverflow.com/q/456672/1132524">class factory</a> does that, but are you sure you actually need that?</p>
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    1. POwhy python exec define class not working
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    1. COwhat I want to do is dynamic create class, in this example the class named 'demo', but the class name is not fixed, it is variable. So my idea is using string to assemble create class statement, the exec the string.
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    2. COWell, then, dynamically create the class, you hardly need `exec` to do that!
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    3. COhow to 'dynamically create the class' ?
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