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POpython dictionary passed as an input to a function acts like a global in that function rather than a local
 

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    <p>I am very confused by the behaviour below. Cases 1, 3, and 4 perform as I would expect, but case 2 does not. Why does case 2 allow the function to change the value of the dictionary entry globally, even though the dictionary is never returned by the function? A main reason I am using functions is to isolate everything in the function from the rest of the code, but this does not seem to be possible if I choose to use the same variable names inside of the function. I was under the understanding that anything explicitly defined in a function is local to that function, but this does not seem to be the case if the dictionary is <strong>defined and passed as an input to the function</strong>.</p> <pre><code>Python 2.7.2+ (default, Oct 4 2011, 20:06:09) [GCC 4.6.1] on linux2 Type "help", "copyright", "credits" or "license" for more information. </code></pre> <p>=============Case 1===============</p> <pre><code>&gt;&gt;&gt; def testfun1(a): ... a=2 ... &gt;&gt;&gt; a=0 &gt;&gt;&gt; testfun1(a) &gt;&gt;&gt; a 0 </code></pre> <p>=============Case 2===============</p> <pre><code>&gt;&gt;&gt; def testfun2(b): ... b['test']=2 ... &gt;&gt;&gt; b={} &gt;&gt;&gt; testfun2(b) &gt;&gt;&gt; b {'test': 2} </code></pre> <p>=============Case 3===============</p> <pre><code>&gt;&gt;&gt; def testfun3(): ... c=2 ... &gt;&gt;&gt; c=0 &gt;&gt;&gt; testfun3() &gt;&gt;&gt; c 0 </code></pre> <p>=============Case 4=============== (explained by this question: <a href="https://stackoverflow.com/questions/14323817/global-dictionaries-dont-need-keyword-global-to-modify-them">Global dictionaries don&#39;t need keyword global to modify them?</a>)</p> <pre><code>&gt;&gt;&gt; def testfun4(): ... d['test']=10 ... &gt;&gt;&gt; d={} &gt;&gt;&gt; testfun4() &gt;&gt;&gt; d {'test': 10} </code></pre>
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