我在二叉树上有一个函数的问题。树中有客户端结构,其中包含一个ID号和一个日期字段。我需要做3个函数,2个find_client函数,一个使用节点的id号进行搜索,一个使用日期,他们都会将地址返回到包含所有匹配节点的新树。在这两个函数上面有一个find_client函数来决定根据用户输入调用哪个函数,我试图使用函数指针来工作,但是我遇到了一个问题。首先,该结构:如何使用函数指针?
typedef struct date{
int day;
int month;
int year;
}date;
typedef struct time{
int hours;
int minutes;
}time;
typedef struct client{
char *first_name;
char *sur_name;
unsigned int id;
unsigned long reg_num;
date rent_date;
time rent_time;
int price_for_day;
}client;
typedef struct client_tree {
struct client c;
struct client_tree *left, *right;
} clientTree;
typedef union clientData{
unsigned int id;
date date;
}clientData;
现在的功能我用:
clientTree* findClient(clientTree* t){
clientTree* n=NULL;
int i=0;
clientData u;
while(i!=1 && i!=2){
printf("\nPlease enter 1 to search via I.D. number, or press 2 to search by date: ");
scanf("%d", &i);
__fpurge(stdin);
if(i==1){
printf("\nEnter the id number please: ");
scanf("%u", &u.id);
__fpurge(stdin);
}
else if (i==2){
printf("\nEnter the day please: ");
scanf("%d", &u.date.day);
__fpurge(stdin);
printf("\nEnter the month please: ");
scanf("%d", &u.date.month);
__fpurge(stdin);
printf("\nEnter the year please: ");
scanf("%d", &u.date.year);
__fpurge(stdin);
}
else
printf("\nNot valid, try again.");
}
clientTree* (*pt2Function)(clientTree*, clientData) = NULL;
pt2Function=GetPtr2(i);
n= (*pt2Function)(t, u);
return n;
}
clientTree* findClientDate(clientTree* t, clientData u){
if(!t)
return NULL;
if(t->c.rent_date.day==u.date.day && t->c.rent_date.month==u.date.month && t->c.rent_date.year==u.date.year){
clientTree* n = createClientTree();
treeAddClient(n,t->c);
n->left=findClientDate(t->left, u);
n->right=findClientDate(t->right, u);
return n;
}
return NULL;
}
clientTree* findClientId(clientTree* t, clientData u){
if(!t)
return NULL;
if(t->c.id==u.id){
clientTree *n = createClientTree();
treeAddClient(n,t->c);
n->left=findClientId(t->left, u);
n->right=findClientId(t->right, u);
return n;
}
return NULL;
}
clientTree*(*GetPtr2(int opCode))(clientTree*, clientData){
if(opCode == 1)
return &findClientId;
else
return &findClientDate;
}
我得到一个错误:“冲突的类型‘GetPtr2’” 我不是很方便用函数指针,有什么建议?
P.S.同时这两种功能要求:
clientTree* treeAddClient(clientTree* root, client c){
if (!root) {
root=createClientTree();
root->c=c;
return root;
}
if (c.id > root->c.id)
root->right = treeAddClient(root->right, c);
else if (c.id < root->c.id)
root->left = treeAddClient(root->left, c);
else
return NULL;
return root;
}
clientTree* createClientTree(){
clientTree *t;
t=ALLOC(clientTree, 1);
return t;
}
刚读函数指针今天这个教程:函数指针教程(http://www.beningo.com/index.php/软件的技术/ 139-一个引入到功能pointers.html)。它可能有帮助。 –