+typedef boost::function<void (string s) > funcptr;
+
+void foo(funcptr fp)
+{
+ fp("hello,world!");
+}
+
+BOOST_PYTHON_MODULE(test)
+{
+ def("foo",foo) ;
+}
+
+
+And then:
+
++>>> def hello(s): +... print s +... +>>> foo(hello) +hello, world! ++ + The short answer is: "you can't". This is not a + Boost.Python limitation so much as a limitation of C++. The + problem is that a Python function is actually data, and the only + way of associating data with a C++ function pointer is to store it + in a static variable of the function. The problem with that is + that you can only associate one piece of data with every C++ + function, and we have no way of compiling a new C++ function + on-the-fly for every Python function you decide to pass + to
foo. In other words, this could work if the C++
+ function is always going to invoke the same Python
+ function, but you probably don't want that.
+
+ If you have the luxury of changing the C++ code you're
+ wrapping, pass it an object instead and call that;
+ the overloaded function call operator will invoke the Python
+ function you pass it behind the object.
+
+
For more perspective on the issue, see this + posting. + +
Revised - 23 January, 2003 + 18 March, 2003