以下代码有什么作用？

Question

可能重复:
你如何阅读C声明？

我不明白以下几点:

int?? ?* (*(*p)[2][2])(int,int);

你能帮我吗？

Answer 1

对于像这样的事情尝试cdecl,解码为;

declare p as pointer to array 2 of array 2 of pointer to function (int, int) returning pointer to int

Answer 2

          p                      -- p
         *p                      -- is a pointer
        (*p)[2]                  -- to a 2-element array
        (*p)[2][2]               -- of 2-element arrays
       *(*p)[2][2]               -- of pointers
      (*(*p)[2][2])(       );    -- to functions  
      (*(*p)[2][2])(int,int);    -- taking 2 int parameters
    * (*(*p)[2][2])(int,int);    -- and returning a pointer
int?? ?* (*(*p)[2][2])(int,int);    -- to int

这种野兽在实践中会是什么样子？

int *q(int x, int y) {...}  // functions returning int *
int *r(int x, int y) {...}
int *s(int x, int y) {...}
int *t(int x, int y) {...}
...
int *(*fptr[2][2])(int,int) = {{p,q},{r,s}};  // 2x2 array of pointers to 
                                              // functions returning int *
...
int *(*(*p)[2][2])(int,int) = &fptr;          // pointer to 2x2 array of pointers
                                              // to functions returning int *
...
int *v0 = (*(*p)[0][0])(x,y);                 // calls q
int *v1 = (*(*p)[0][1])(x,y);                 // calls r
... // etc.