33.函数指针相关问题

函数指针作为参数

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <Windows.h>
 4 
 5 int add(int a, int b)
 6 {
 7     return a + b;
 8 }
 9 
10 int sub(int a, int b)
11 {
12     return a - b;
13 }
14 
15 int mul(int a, int b)
16 {
17     return a*b;
18 }
19 
20 int chu(int a, int b)
21 {
22     return a / b;
23 }
24 
25 //函数指针作为参数
26 int run(int(*p)(int,int), int a, int b)
27 {
28     return p(a, b);
29 }
30 
31 void main()
32 {
33     int res = run(add, 10, 2);
34     printf("%d
", res);
35     system("pause");
36 }

函数指针作为返回值

1 //返回一个函数指针
2 //返回的类型是(*p)(int,int)
3 int (*p())(int,int)
4 {
5     return add;
6 }

一个复杂的例子，返回值是函数指针，参数是函数指针

 1 //返回一个函数指针
 2 //返回的类型是(*p)(int,int)
 3 int(*p(  int(*op)(int,int)  ))(int, int)
 4 {
 5     return op;
 6 }
 7 
 8 void main()
 9 {
10     printf("%d
", p(sub)(1,2));
11     system("pause");

12 }

typedef 简化函数指针

1 typedef int(*pf)(int, int);
2 pf p1 = add;
3 printf("%d", p1(1, 2));

函数指针数组

1     int(**p)(int, int) = (int (*[])(int, int)){ add, sub, mul, chu };
2     printf("%d
",(*(p+1))(1, 2));
3     system("pause");

typedef简化函数指针数组

1 typedef int(**pp)(int, int);
2 typedef int(*px[10])(int, int);

1 pp pp1 = (px){ add, sub };
2 printf("%d
", pp1[0](2,4));