20200708 千锤百炼软工人第三天
Posted huangmouren233
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了20200708 千锤百炼软工人第三天相关的知识,希望对你有一定的参考价值。
今天是第三天
今天的任务完成不理想
今天的任务是完成一个通讯录程序
其储存方式就是需要应用链表来实现
其主要功能有
创建,显示,修改,插入,删除,文件的导入,文件的导出。
其中前五个功能基本完成,后两个功能还没实现。
但是前五个功能的程序设计基本完成但是具体实现阶段不理想
花费了很长时间来改错,但是最终虽然没有了错误但是实现起来还是有所偏差
#include<iostream>
#include<stdlib.h>
#include<string>
using namespace std;
const int MAX_N = 100;
typedef struct user{
unsigned m_id;
string m_name;
unsigned m_age;
string m_sex;
long m_Hphone;
long m_phone;
}user;
typedef class User
{
public:
unsigned m_id;
string m_name;
unsigned m_age;
string m_sex;
long m_Hphone;
long m_phone;
class User *next;
}ListNode;
class List
{
public:
#include<stdlib.h>
#include<string>
using namespace std;
const int MAX_N = 100;
typedef struct user{
unsigned m_id;
string m_name;
unsigned m_age;
string m_sex;
long m_Hphone;
long m_phone;
}user;
typedef class User
{
public:
unsigned m_id;
string m_name;
unsigned m_age;
string m_sex;
long m_Hphone;
long m_phone;
class User *next;
}ListNode;
class List
{
public:
List(struct user *a, int n)
{
ListNode *p, *s;
int i;
L = (ListNode *)malloc(sizeof(ListNode));
p = L;
for (i = 0; i < n; i++)
{
s = (ListNode *)malloc(sizeof(ListNode));
s->m_id = a[i].m_id;
s->m_name = a[i].m_name;
s->m_age = a[i].m_age;
s->m_sex = a[i].m_sex;
s->m_Hphone = a[i].m_Hphone;
s->m_phone = a[i].m_phone;
p->next = s;
p = s;
cout<<"建立完成!"<<endl;
}
p->next = NULL;
cout<<"建立完成!3"<<endl;
{
ListNode *p, *s;
int i;
L = (ListNode *)malloc(sizeof(ListNode));
p = L;
for (i = 0; i < n; i++)
{
s = (ListNode *)malloc(sizeof(ListNode));
s->m_id = a[i].m_id;
s->m_name = a[i].m_name;
s->m_age = a[i].m_age;
s->m_sex = a[i].m_sex;
s->m_Hphone = a[i].m_Hphone;
s->m_phone = a[i].m_phone;
p->next = s;
p = s;
cout<<"建立完成!"<<endl;
}
p->next = NULL;
cout<<"建立完成!3"<<endl;
}
~List(){
ListNode *pre = L, *p = L->next;
while (p->next != NULL)
{
free(pre);
pre = p;
p = pre->next;
}
free(pre);
}
bool isEmpty()
{
return (L->next == NULL);
}
int getLength()
{
ListNode *p = L;
int i = 0;
while (p->next != NULL)
{
i++;
p = p->next;
}
return i;
}
void DisplayList(){
ListNode *p = L->next;
while (p != NULL){
cout << p->m_id<<" "<<p->m_name<<" "<<p->m_age<<" "<<p->m_sex<<" "<<p->m_Hphone<<" "<<p->m_phone<<endl;
p = p->next;
}
cout << endl;
}
bool getElem(string &N, unsigned &id,unsigned &age,string &name,string &sex,long &Hphone,long &phone){
ListNode *p = L;
while (N.compare(p->m_name)&&p->next != NULL){
p = p->next; }
if (N.compare(p->m_name)&&p->next == NULL){
return false;
}
else{
id = p->m_id;
age=p->m_age;
name=p->m_name;
sex=p->m_sex;
Hphone=p->m_Hphone;
phone=p->m_phone;
return true;
}
}
int LocateElem(long &elem){
ListNode *p = L;
int i = 0;
while (elem == p->m_phone&&p->next != NULL){
i++;
p = p->next;
}
if (elem==p->m_phone&&p->next == NULL){
return 0;
}
else{
return i;
}
}
private:
ListNode *L;
};
int main()
{
~List(){
ListNode *pre = L, *p = L->next;
while (p->next != NULL)
{
free(pre);
pre = p;
p = pre->next;
}
free(pre);
}
bool isEmpty()
{
return (L->next == NULL);
}
int getLength()
{
ListNode *p = L;
int i = 0;
while (p->next != NULL)
{
i++;
p = p->next;
}
return i;
}
void DisplayList(){
ListNode *p = L->next;
while (p != NULL){
cout << p->m_id<<" "<<p->m_name<<" "<<p->m_age<<" "<<p->m_sex<<" "<<p->m_Hphone<<" "<<p->m_phone<<endl;
p = p->next;
}
cout << endl;
}
bool getElem(string &N, unsigned &id,unsigned &age,string &name,string &sex,long &Hphone,long &phone){
ListNode *p = L;
while (N.compare(p->m_name)&&p->next != NULL){
p = p->next; }
if (N.compare(p->m_name)&&p->next == NULL){
return false;
}
else{
id = p->m_id;
age=p->m_age;
name=p->m_name;
sex=p->m_sex;
Hphone=p->m_Hphone;
phone=p->m_phone;
return true;
}
}
int LocateElem(long &elem){
ListNode *p = L;
int i = 0;
while (elem == p->m_phone&&p->next != NULL){
i++;
p = p->next;
}
if (elem==p->m_phone&&p->next == NULL){
return 0;
}
else{
return i;
}
}
private:
ListNode *L;
};
int main()
{
ListNode *m_node;
int n,i;
user a[MAX_N];
string name,sex,N;
long Hphone,phone,elem;
string q;
unsigned age,id;
cout << "Input the length:" <<endl;
cin >> n;
for (i = 0; i < n; i++)
{
cin >> a[i].m_id;
cin >> a[i].m_name;
cin >> a[i].m_age;
cin >> a[i].m_sex;
cin >> a[i].m_Hphone;
cin >> a[i].m_phone;
cout<<"下一位用户信息:"<<endl;
}
List m_list(a, n);
m_list.DisplayList();
if (!m_list.isEmpty())
{
cout << "it is not empty." << endl;
cout << "The length:" << m_list.getLength() << endl;
}
else{
cout << "It is empty." << endl;
}
cout << "Input number:" << ends;
cin >> q;
if (m_list.getElem(q,id,age,name,sex,Hphone,phone))
{
cout << "NO. " << id <<" ,the name is"<<name<<" ,the age is"<<age<<" ,the sex is"<<sex<<" ,the Hphone is"<<Hphone<<" ,the phone is"<<phone<<endl;
}
else
{
cout << "error number or not contained." << endl;
}
cout << "Input the data:" << ends;
cin >> elem;
if (m_list.LocateElem(elem) == 0)
{
cout << "Not contained." << endl;
}
else
{
cout << "The location is No. " << m_list.LocateElem(elem) << endl;
}
}
int n,i;
user a[MAX_N];
string name,sex,N;
long Hphone,phone,elem;
string q;
unsigned age,id;
cout << "Input the length:" <<endl;
cin >> n;
for (i = 0; i < n; i++)
{
cin >> a[i].m_id;
cin >> a[i].m_name;
cin >> a[i].m_age;
cin >> a[i].m_sex;
cin >> a[i].m_Hphone;
cin >> a[i].m_phone;
cout<<"下一位用户信息:"<<endl;
}
List m_list(a, n);
m_list.DisplayList();
if (!m_list.isEmpty())
{
cout << "it is not empty." << endl;
cout << "The length:" << m_list.getLength() << endl;
}
else{
cout << "It is empty." << endl;
}
cout << "Input number:" << ends;
cin >> q;
if (m_list.getElem(q,id,age,name,sex,Hphone,phone))
{
cout << "NO. " << id <<" ,the name is"<<name<<" ,the age is"<<age<<" ,the sex is"<<sex<<" ,the Hphone is"<<Hphone<<" ,the phone is"<<phone<<endl;
}
else
{
cout << "error number or not contained." << endl;
}
cout << "Input the data:" << ends;
cin >> elem;
if (m_list.LocateElem(elem) == 0)
{
cout << "Not contained." << endl;
}
else
{
cout << "The location is No. " << m_list.LocateElem(elem) << endl;
}
}
具体的代码在此奉上
希望有位大佬可以稍微指点一二
今天的java学习稍微落后一点
学习的比较少
所以也就没办法在此详细记录
明天继续完成小学期任务,并赶上Java学习的进度
以上是关于20200708 千锤百炼软工人第三天的主要内容,如果未能解决你的问题,请参考以下文章