所以a是个5个元素的数组,每个元素为函数指针.

可用代码验证：

#include 
    
     #include 
     
      int* (*a[5])(int,char*);int *foo(int n, char *s){	int *p;	p = (int *)malloc(sizeof(int));	*p = n + atoi(s);	return p;}int main(int argc, char *argv[]){	int *p;	a[0] = &foo;	p = (*a[0])(1, "2");	printf("%d\n", *p);	return 0;}
     
    

输出：

3

顺便把函数指针再复习一下：

#include 
    
     #include 
     
      //int* (*a[5])(int,char*);typedef int (*f)(int,char*);f funp;int foo(int n, char *s){	return 10086;}int main(int argc, char *argv[]){	funp = &foo;	printf("%x\n",funp);	printf("%d\n",(*funp)(6,"test"));	return 0;}
     
    

3f1000