c语言的函数声明和typedef - fun - fun
c语言的函数声明和typedef
abelard
posted @ 2011年5月26日 08:55
in C语言
, 2266 阅读
在《UNIX环境高级编程》、《UNIX® Network Programming Volume 1, Third Edition: The Sockets Networking API》都讨论信号处理,并且都解释了下面这个函数原型:
void (*signal (int signo, void (*func) (int))) (int);
没搞懂,只好找到《c 陷阱与缺陷》对函数定义的解释,初步的理解是:
1. 函数的返回类型可以是基本类型和复合类型,还可以是函数.上面的原型,就属于返回类型为函数:
void (*func)(int);
2. 如果使用typedef定义上面的函数,看起来是要舒服多了,但是理解起来是一样的:
typedef void (*func)(int);
func signal(int, func pFunc);
(注:感觉理解的还很肤浅)
做两个例子来看看到底是怎么用的:
#include <stdio.h> typedef void (*signal_handler)(int); void (*(signal_hand(int, void sig(int))))(int); signal_handler pSig_Handle(int , signal_handler); signal_handler pSig_Handle(int a, signal_handler p) { printf("1111 %d \n",a); return(p); } void func(int a){printf("2222 %d\n",a);} void (*(signal_hand(int a, void sig(int x))))(int b){ printf("3333 %d \n",a); return(sig); } main() { signal_handler psigHandler = pSig_Handle(50,func); signal_handler psigHand = signal_hand(200,func); psigHandler(100); psigHand(300); signal_hand(200,func)(5000); }
输出的结果是:
$ ./a.out 1111 50 3333 200 2222 100 2222 300 3333 200 2222 5000
2022年8月06日 22:12
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