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    <blockquote> <p>So in a function the adding a 6 to the set doesn't show but it does when not in a function?</p> </blockquote> <p>No, that is not what happens.</p> <p>What happens is that, when you execute <code>mylist = mylist + [6]</code>, you are effectively creating an entirely new list and putting it in the local <code>mylist</code> variable. This <code>mylist</code> variable will vanish after the execution of the function and the newly created list will vanish as well.</p> <p>OTOH when you execute <code>mylist.append(6)</code> you do not create a new list. You get the list already in the <code>mylist</code> variable and add a new element to this same list. The result is that the list (which is pointed by <code>list2</code> too) will be altered itself. The <code>mylist</code> variable will vanish again, but in tis case you altered the original list.</p> <p>Let us see if a more visual explanation can help you :)</p> <h1>What happens when you call <code>proc()</code></h1> <p>When you write <code>list1 = [1, 2, 3, 4, 5]</code> you are creating a new list object (at the right side of the equals sign) and creating a new variable, <code>list1</code>, which will point to this object.</p> <p><img src="https://i.imgur.com/7Wifq.png" alt="Creating new list instance and global variable"></p> <p>Then, when you call <code>proc()</code>, you create another new variable, <code>mylist</code>, and since you pass <code>list1</code> as parameter, <code>mylist</code> will point to the same object:</p> <p><img src="https://i.imgur.com/KowHP.png?1" alt="Calling method creates local variable"></p> <p>However, the operation <code>mylist + [6]</code> <em>creates a whole new list object</em> whose contents are the contents of the object pointed by <code>mylist</code> plus the content of the following list object - that is, <code>[6]</code>. Since you attribute this new object to <code>mylist</code>, our scenario changes a bit and <code>mylist</code> does not point to the same object pointed by <code>list1</code> anymore:</p> <p><img src="https://i.imgur.com/X6yaB.png?1" alt="mylist points to new list object"></p> <p>What I have not said is that <code>mylist</code> is a <em>local variable</em>: it will disappear after the end of the <code>proc()</code> function. So, when the <code>proc()</code> execution ended, the <code>mylist</code> is gone:</p> <p><img src="https://i.imgur.com/8NP33.png?1" alt="mylist is gone"></p> <p>Since no other variable points to the object generated by <code>mylist + [6]</code>, it will disappear, too (since the garbage collector* will collect it):</p> <p><img src="https://i.imgur.com/ddifs.png?1" alt="GC collects the list"></p> <p>Note that, in the end, the object pointed by <code>list1</code> is not changed.</p> <h1>What happens when you call <code>proc2()</code></h1> <p>Everything changes when you call <code>proc2()</code>. At first, it is the same thing: you create a list...</p> <p><img src="https://i.imgur.com/7Wifq.png" alt="Creating new list instance and global variable"></p> <p>...and pass it as a parameter to a function, which will generate a local variable:</p> <p><img src="https://i.imgur.com/KowHP.png?1" alt="Calling method creates local variable"></p> <p>However, instead of using the <code>+</code> concatenation operator, which generates a new list, you apply the <code>append()</code> method to the existing list. The <code>append()</code> method <em>does not create a new object</em>; instead, it _changes the existing one:</p> <p><img src="https://i.imgur.com/rv7Br.png?1" alt="Appending a value to a list"></p> <p>After the end of the function, the local variable will disappear, but the original object pointed by it and by <code>list1</code> will be already altered:</p> <p><img src="https://i.imgur.com/VDnSF.png?1" alt="It is still altered"></p> <p>Since it is still pointed by <code>list1</code>, the original list is not destroyed.</p> <p><strong>EDIT</strong>: if you want to take a look at all this stuff happening <strong>before your eyes</strong> just go to <a href="http://www.pythontutor.com/visualize.html#code=list1+=+%5B1,2,3,4%5D%0Alist2+=+%5B1,2,3,4%5D%0A%0Adef+proc%28mylist%29:%0A++++mylist+=+mylist+++%5B6%5D%0A%0Adef+proc2%28mylist%29:%0A++++mylist.append%286%29%0A%0A#+Can+you+explain+the+results+given+by+the+four+print+statements+below?%2bRemove%0A#%2bthe%2bhashes%2b#%2band%2brun%2bthe%2bcode%2bto%2bcheck.%0A%0Aprint%2b%28list1%29%0Aproc%28list1%29%0Aprint%2b%28list1%29%0A%0Aprint%2b%28list2%29%0Aproc2%28list2%29%0Aprint%2b%28list2%29&amp;mode=display&amp;origin=opt-frontend.js&amp;cumulative=false&amp;heapPrimitives=false&amp;drawParentPointers=false&amp;textReferences=false&amp;showOnlyOutputs=false&amp;py=2&amp;rawInputLstJSON=%5B%5D&amp;curInstr=16" rel="nofollow noreferrer">this radically amazing simulator</a>:</p> <p><img src="https://i.stack.imgur.com/4p5dC.gif" alt="enter image description here"></p> <p>* If you do not know what is garbage collector... well, you will discover soon after understanding your own question.</p>
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    1. COThere is also good [service](http://people.csail.mit.edu/pgbovine/python/tutor.html#mode=edit) to go python code step by step and look at all variables in browser! Very close to what you wrote, @brandizzi
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    2. CO+1 for so great drawing ..
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    3. CO@stalk I did not know this service. it is just great! Thanks!
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