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POHow Does Calling Work In Python?
 

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  1. POHow Does Calling Work In Python?
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    <p>For a project I'm working on, I'm implementing a linked-list data-structure, which is based on the idea of a pair, which I define as:</p> <pre><code>class Pair: def __init__(self, name, prefs, score): self.name = name self.score = score self.preferences = prefs self.next_pair = 0 self.prev_pair = 0 </code></pre> <p>where <code>self.next_pair</code> and <code>self.prev_pair</code> are pointers to the previous and next links, respectively. </p> <p>To set up the linked-list, I have an install function that looks like this.</p> <pre><code>def install(i, pair): flag = 0 try: old_pair = pair_array[i] while old_pair.next_pair != 0: if old_pair == pair: #if pair in remainders: remainders.remove(pair) return 0 if old_pair.score &lt; pair.score: flag = 1 if old_pair.prev_pair == 0: # we are at the beginning old_pair.prev_pair = pair pair.next_pair = old_pair pair_array[i] = pair break else: # we are not at the beginning pair.prev_pair = old_pair.prev_pair pair.next_pair = old_pair old_pair.prev_pair = pair pair.prev_pair.next_pair = pair break else: old_pair = old_pair.next_pair if flag==0: if old_pair == pair: #if pair in remainders: remainders.remove(pair) return 0 if old_pair.score &lt; pair.score: if old_pair.prev_pair==0: old_pair.prev_pair = pair pair.next_pair = old_pair pair_array[i] = pair else: pair.prev_pair = old_pair.prev_pair pair.next_pair = old_pair old_pair.prev_pair = pair pair.prev_pair.next_pair = pair else: old_pair.next_pair = pair pair.prev_pair = old_pair except KeyError: pair_array[i] = pair pair.prev_pair = 0 pair.next_pair = 0 </code></pre> <p>Over the course of the program, I am building up a dictionary of these linked-lists, and taking links off of some and adding them in others. Between being pruned and re-installed, the links are stored in an intermediate array.</p> <p>Over the course of debugging this program, I have come to realize that my understanding of the way Python passes arguments to functions is flawed. Consider this test case I wrote:</p> <pre><code>def test_install(): p = Pair(20000, [3, 1, 2, 50], 45) print p.next_pair print p.prev_pair parse_and_get(g) first_run() rat = len(juggler_array)/len(circuit_array) pref_size = get_pref_size() print pref_size print install(3, p) print p.next_pair.name print p.prev_pair </code></pre> <p>When I run this test, I get the following result.</p> <pre><code>0 0 10 None 10108 0 </code></pre> <p>What I don't understand is why the second call to <code>p.next_pair</code> produces a different result (<code>10108</code>) than the first call (<code>0</code>). <code>install</code> does not return a <code>Pair</code> object that can overwrite the one passed in (it returns <code>None</code>), and it's not as though I'm passing <code>install</code> a pointer.</p> <p>My understanding of call-by-value is that the interpreter copies the values passed into a function, leaving the caller's variables unchanged. For example, if I say</p> <pre><code>def foo(x): x = x+1 return x baz = 2 y = foo(baz) print y print baz </code></pre> <p>Then <code>3</code> and <code>2</code> should be printed, respectively. And indeed, when I test that out in the Python interpreter, that's what happens.</p> <p>I'd really appreciate it if anyone can point me in the right direction here. </p>
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    1. COPython doesn't have call-by-value. Nor does it have call-by-reference.
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      1. USIgnacio Vazquez-Abrams
    2. COFYI, you can use the `{}` button at the top of the edit box to indent a selected code block by 4 spaces -- it's an easy way to maintain correct indentation.
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    3. COThat `Pair` class looks like it should probably be a `dict` instead.
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      1. USDaenyth
 

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