关于“ void（* fcn）（void *）”类型的参数的c函数声明说明

Question

I encountered the following line as a C function declaration. I'm confused by the type of the first argument passed in. Can anyone explain how to understand the type of the first argument?

int clone(void (*fcn) (void *) , void *arg, void *stack)

Answer 1

void (*fcn) (void *)

从内到外读取类型： fcn是指向函数的指针，该函数采用void *参数，但不返回任何内容（ void ）。

Answer 2

Use "Spiral Rule" :

      +------+
      | +--+ |
      | ^  | |
void (*fcn ) (void *)
  ^   ^    | |
  |   +----+ |
  +----------+

So,

• fcn is a pointer to
• a function having void * as an argument
• returning void (nothing)

Answer 3

After a few editing (to remove the names of variables) cdecl gave the answer :

declare clone as function (pointer to function (pointer to void) returning void, pointer to void, pointer to void) returning int

Or, you can test only the first argument and get it's type: pointer to function of pointer to void, returning void.