我们如何从被调用者修改调用者中的值

Question

如何从被调用者修改调用者中的值？

在下面的代码中，为什么exec不修改函数bar中变量xxxx的值？

def foo(xcode):
print("CODE: ", xcode)
of = inspect.currentframe().f_back
of_locals = of.f_locals
of_globals = inspect.stack()[1][0].f_globals
exec(xcode, of_globals, of_locals)
del of

def bar():
x = 1
y = 2
z = 3
foo("x = y + z")
print("x == ", x)
print("x == 6? ", x == 6)
bar()

Answer 1

在exec语句后添加以下行似乎有效：

ctypes.pythonapi.PyFrame_LocalsToFast(ctypes.py_object(of), ctypes.c_int(0))

因此，我们有：

import ctypes
def foo(xcode):
print("CODE: ", xcode)
of = inspect.currentframe().f_back
of_locals = of.f_locals
of_globals = inspect.stack()[1][0].f_globals
exec(xcode, of_globals, of_locals)
ctypes.pythonapi.PyFrame_LocalsToFast(ctypes.py_object(of), ctypes.c_int(0))
del of