Question

如果你有int myfunc(struct mystruct **context)这样的函数，那么在C中 OCaml外国签名将是

    type mystruct
    let mystruct : mystruct structure typ = structure "mystruct"

    let myfunc = foreign "myfunc" (ptr (ptr mystruct) @-> returning int)

问题是如何调用该函数？我们不能addr (addr mystructinstance)，所以代码不会编译：

    let inst = make mystruct in
    let res = myfunc (addr (addr inst))

Answer 1

ctypes的内存模型与C紧密匹配，因此您需要编写的代码将与您在C中编写的内容相匹配。您对addr (addr mystructinstance)生病的类型是正确的，即因为它对应于&(&mystructinstance)，这在C中没有任何意义。

在C中，您可能会写出类似的内容：

mystruct *p = NULL;
int err = myfunc(&p);
if (err == ok) {
  // p has now a usable value for the rest of the API
}

ctypes中的等效代码将是

let p_addr = Ctypes.allocate (ptr mystruct) (from_voidp mystruct null) in
let err = myfunc p_addr in
(* check err *)
let p = !@ p_addr in
(* do things with p, typed mystruct structure ptr *)

OCaml Ctypes - 如何使指针指向

1 个答案: