指向结构函数指针的函数指针

Question

Today I received a homework, about function pointer(binary tree) . There is a chunk of code that i can not understand... I understand what a function pointer is and how it work but what about 2 function pointers (1 is in a parameter )

header.h

typedef struct _node_ {
int key;
struct _node_ *left;
struct _node_ *right;
} node;

typedef struct _bstree_ {
node *root_node;
int (*compare_keys)(int x, int y);  //this code
} bstree;

c file

void bst_init_with_comp_operator(bstree *bst, int(*comp)(int x, int y)) {
bst_init(bst);

bst->compare_keys = comp; // what does this mean? a fucntion pointer parameter to function pointer to struct?

 }

// comp is supposed to return -1 if x should be treated as less than y,zero if x should enter code here` be treated as equal to y, or 1 if x should be treated as greater than y.

so I created this function:

int compare (int x , int y)
{
if(x < y) return -1;
else if(x == y) return 0;
else return 1;

}

main.c

bstree tree;
bst_init_with_comp_operator(&tree,compare(2,3))

but it doesnt work ...

normally we just need something like this

 int function(int x, int y)
 { return x+y;}

 int (*pointerf) (int , int)

 pointerf = function;

 pointerf(2,3)

Answer 1

您只需要bst_init_with_comp_operator(&tree,compare)) ，这将传递一个函数地址作为参数，而如果您进行compare(a,b)则传递该函数的结果，因此整数1、0或-1。