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    <p>The catch-all <code>*args</code> parameter <em>must</em> come after any explicit arguments:</p> <pre><code>def foo(func, named_arg=None, *args): </code></pre> <p>If you also add the catch-all <code>**kw</code> keywords parameter to a definition, then that <em>has</em> to come after the <code>*args</code> parameter:</p> <pre><code>def foo(func, named_arg=None, *args, **kw): </code></pre> <p>Mixing explicit keyword arguments and the catch-all <code>*args</code> argument does lead to unexpected behaviour; you cannot both use arbitrary positional arguments <em>and</em> explicitly name the keyword arguments you listed at the same time.</p> <p>Any extra positionals beyond <code>func</code> are first used for <code>named_arg</code> which can also act as a positional argument:</p> <pre><code>&gt;&gt;&gt; def foo(func, named_arg = None, *args): ... print func, named_arg, args ... &gt;&gt;&gt; foo(1, 2) 1 2 () &gt;&gt;&gt; foo(1, named_arg=2) 1 2 () &gt;&gt;&gt; foo(1, 3, named_arg=2) Traceback (most recent call last): File "&lt;stdin&gt;", line 1, in &lt;module&gt; TypeError: foo() got multiple values for keyword argument 'named_arg' &gt;&gt;&gt; foo(1, 2, 3) 1 2 (3,) </code></pre> <p>This is because the any second positional argument to <code>foo()</code> will <em>always</em> be used for <code>named_arg</code>.</p> <p>In Python 3, the <code>*args</code> parameter <em>can</em> be placed before the keyword arguments, but that has a <em>new</em> meaning. Normally, keyword parameters can be specified in the call signature as positional arguments (e.g. call your function as <code>foo(somefunc, 'argument')</code> would have <code>'argument'</code> assigned to <code>named_arg</code>). By placing <code>*args</code> or a plain <code>*</code> in between the positional and the named arguments you exclude the named arguments from being used as positionals; calling <code>foo(somefunc, 'argument')</code> would raise an exception instead.</p>
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    1. COwhen I do it like that and define a function as follows `def func(text): print "Some text: ", text` and call it via `foo` as `foo(func,'world')` I get the error `TypeError: func() takes exactly 1 argument (0 given)`. And a call `foo(func, named_arg = 1, "world")` also does not work (`SyntaxError: non-keyword arg after keyword arg`).
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    2. CO@Alex: `foo(func, 1, 'world')` does work.
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    3. COyes this does. thanks for the explanation.
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