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    <p>In some ways, all calls in Python are called with references. In fact, all variables are references in a sense. But some types, like <code>int</code> from your example, cannot be changed.</p> <p>In the case of, say, a <code>list</code>, the functionality you're looking for is trivial:</p> <pre><code>def change_it(some_list): some_list.append("world") foo = ["hello"] change_it(foo) print(foo) # prints ['hello', 'world'] </code></pre> <p>Note, however, that reassigning the <em>parameter</em> variable <code>some_list</code> does not change the value in the calling context. </p> <p>If you're asking this question, though, you're probably looking to do something like set two or three variables using one function. In that case, you're looking for something like this:</p> <pre><code>def foo_bar(x, y, z): return 2*x, 3*y, 4*z x = 3 y = 4 z = 5 x, y, z = foo_bar(x, y, z) print(y) # prints 12 </code></pre> <p>Of course, you can do anything in Python, but that doesn't mean you should. In the fashion of the TV show Mythbusters, here's something that does what you're looking for</p> <pre><code>import inspect def foo(bar): frame = inspect.currentframe() outer = inspect.getouterframes(frame)[1][0] outer.f_locals[bar] = 2 * outer.f_locals[bar] a = 15 foo("a") print(a) # prints 30 </code></pre> <p>or even worse:</p> <pre><code>import inspect import re def foo(bar): # get the current call stack my_stack = inspect.stack() # get the outer frame object off of the stack outer = my_stack[1][0] # get the calling line of code; see the inspect module documentation # only works if the call is not split across multiple lines of code calling_line = my_stack[1][4][0] # get this function's name my_name = my_stack[0][3] # do a regular expression search for the function call in traditional form # and extract the name of the first parameter m = re.search(my_name + "\s*\(\s*(\w+)\s*\)", calling_line) if m: # finally, set the variable in the outer context outer.f_locals[m.group(1)] = 2 * outer.f_locals[m.group(1)] else: raise TypeError("Non-traditional function call. Why don't you just" " give up on pass-by-reference already?") # now this works like you would expect a = 15 foo(a) print(a) # but then this doesn't work: baz = foo_bar baz(a) # raises TypeError # and this *really*, disastrously doesn't work a, b = 15, 20 foo_bar, baz = str, foo_bar baz(b) and foo_bar(a) print(a, b) # prints 30, 20 </code></pre> <p>Please, <em>please</em>, <em><strong>please</strong></em>, don't do this. I only put it in here to inspire the reader to look into some of the more obscure parts of Python.</p>
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    1. POSend by ref/by ptr in python?
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    1. USDavid Hollman
    1. USDavid Hollman
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    1. CONot call by reference, call be reference objects. All parameters are passed by reference by default.
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      1. USGames Brainiac
    2. COGood point. Fixing it now
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      1. USDavid Hollman
    3. COYour examples using `f_locals` doesn't work in the general case. whenever the given name is a fast local (as it is for variables in local scope of most functions), then the `f_locals` dict won't contain the value, it's in `f_localsplus`, which is not accessible from normal python. There's no reasonable way to push data from `f_locals` into `f_localsplus` (except with ctypes).
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      1. USSingleNegationElimination
 

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