CFFI 将回调作为参数传递给另一个函数

Question

这是我的误解的解释：

我有一个名为 Codex 的类：

class Codex:
    ffi = FFI()
    def __init__(self):
        self.ffi = FFI()
        self.ClibDec = self.ffi.dlopen("...")
        self.ffi.cdef("""int CRD_Set(int handle, int property, void *value);
                         int CRD_Get(int handle, int property, void *value);
                      """)

    def funct1(self):
        handle = self.ClibDec."something"
        self.ClibDec.CRD_Set(handle, 0, funct2)

    @ffi.callback("int(*)(int, int)")
    def funct2(self, handle):

        status = 0
        self.ClibDec.CRD_Get(handle, 420, status)
        # do something

        return 0

我的问题是如何在 funct1 的 self.ClibDec.CRD_Set 内部调用 funct2 ？

我检查了几个类似的问题和文档： https://cffi.readthedocs.io/en/latest/ using.html#callbacks-old-style

但我还是不明白callback的用法（旧风格）

Answer 1

也许您正在寻找这样的代码？这将是最接近您要编写的解决方案。

class MyClass:

    def funct1(self):
        handle = self.ClibDec."something"
        # make a cdata function pointer object that will call
        # back self.funct2(handle)
        cfnptr = ffi.callback("int(*)(int)", self.funct2)
        self.ClibDec.CRD_Set(handle, 0, cfnptr)
        # the cfnptr pointer must not be called after
        # the variable 'cfnptr' goes out of scope,
        # which occurs when funct1() returns here.

    def funct2(self, handle):
        ...
        return 0