Have you tried passing in empty dicts for globals and locals? I think that the defaults will be the *current* globals and locals, and then of course your namespace is broken...
On Tue, Oct 7, 2008 at 1:26 PM, Gordon Fraser <[EMAIL PROTECTED]> wrote: > Hi, > > I'm trying to parse Python code to an AST, apply some changes to the AST > and then compile and run the AST, but I'm running into problems when > trying to evaluate/execute the resulting code object. It seems that the > global namespace differs depending on where I call parse and eval/exec. > > The following code parses a file, compiles and then evaluates the AST. > If I call Python directly on this code, then it works: > > import sys, parser > ast = parser.suite(open(sys.argv[1]).read()) > code = ast.compile() > exec(code) > > ...and it also works this way with Python2.6: > > ast = compile(open(sys.argv[1]).read(), "<AST>", > 'exec',_ast.PyCF_ONLY_AST) > code = compile(ast, "<AST", "exec") > exec(code) > > However, if I include that snippet in a different scope (some function > or class), then the namespace that the code object will have differs - > it seems the symbols defined in the AST are not included when executing > the code. For example: > > import sys,parser > > def main(): > ast = parser.suite(open(sys.argv[1]).read()) > code = ast.compile() > exec(code) > > if __name__ == "__main__": > main() > > In particular this is a problem if I'm parsing a module with several > functions - none of these functions actually ends up in the scope of the > code object (same behavior with Python2.6 and the PyCF_ONLY_AST > version). > > The function "exec" takes parameters for globals and locals, but I have > no idea where to get these dictionaries from the "parser" module. My > guess is that I am misunderstanding something about how Python treats > namespaces. Can anyone help me here? > > Thanks, > Gordon > > > -- > http://mail.python.org/mailman/listinfo/python-list > -- [EMAIL PROTECTED] http://orestis.gr -- http://mail.python.org/mailman/listinfo/python-list
