C++ - 结构体指针数组

Question

C++ - 结构体指针数组

3

我有一个小程序，其中有两个结构体： Person - 包含id和基本方法 Ppl - 包含一个人员数组以及一些用于操作该数组的方法。

struct Person {
 const int id;
 Person();
 Person(int a);
};

Person::Person(int a) : id(a) {}

这是Person结构体及其方法。

const int MAX = 5;

设置数组长度限制

struct Ppl {
 static int current;   //Keeps track of current position in array
 Person *ppls[MAX];    //Creates an array of Person structure pointers
 void add(int a);      //Adds a person with id to the next available position
 //void remove(int b);
 int searchid(int c);  //Searches ppls for an id c.
 Ppl();                //Constructor
};

int Ppl::current = 0;  //Initializing static variable

void Ppl::add(int a) {
 int ret = searchid(a);          //Determine if id a already exists in ppls
 //cout << "Ret: " << ret << endl;
 if (ret > MAX) {                //If value returned is > MAX, value exists
  cout << "User " << a << " already exists" << endl;
 } else {                        //else, add it to the next available spot
  Person p(a);
  ppls[current] = &p;
  cout << "Added user: " << a << " at index: " << current << endl;
  current++;
 }
}

Ppl::Ppl() {
 current = 0;
 int i = 0;
 while (i < MAX) {    //Sets all Person pointers to NULL on initialization
  ppls[i] = NULL;
  cout << "ppls[" << i << "] is NULL" << endl;
  i++;
 }
}

int Ppl::searchid(int c) {
 int i = 0;
 bool found = false;
 while(i < MAX) {
  if (ppls[i] == NULL) {  //If NULL, then c wasn't found in array because
   break;                 //there is no NULL between available spots.
  } else {
    if (ppls[i]->id == c) {
     found = true;
   }
  }
  i++;
 }
 if (found == true) {
  return 10;     //A number higher than MAX
 } else {
  return 1;      //A number lower than MAX
 }
}

主要功能是:

int main() {
 Ppl people;
 people.add(21);
 cout << people.ppls[0]->id << endl;
 people.add(7);
 cout << people.ppls[0]->id << " ";
 cout << people.ppls[1]->id << endl;
 people.add(42);
 cout << people.ppls[0]->id << " ";
 cout << people.ppls[1]->id << " ";
 cout << people.ppls[2]->id << endl;
 people.add(21);
 cout << people.ppls[0]->id << " ";
 cout << people.ppls[1]->id << " ";
 cout << people.ppls[2]->id << " ";
 cout << people.ppls[3]->id << endl;
}

我得到的输出是：

ppls[0] is NULL
ppls[1] is NULL
ppls[2] is NULL
ppls[3] is NULL
ppls[4] is NULL
Added user: 21 at index: 0
21
Added user: 7 at index: 1
7 0
Added user: 42 at index: 2
42 0 0
Added user: 21 at index: 3
21 0 0 0

为什么它会将所有新条目添加到数组的开头，同时保持其余部分为空？为什么它没有检测到21已经被添加。

我一直在努力想弄清楚这个问题。非常感谢任何帮助。感谢社区。

编辑我已经修复了它，使它将元素添加到数组中并识别出id是否存在。

我通过添加析构函数对Ppl结构进行了更改：

Ppl::~Ppl() {
 int i = 0;
 while (i < MAX) {
  delete ppls[i];
  i++;
 }
}

通过修改 add 方法。

void Ppl::add(int a) {
 int ret = searchid(a);
 //cout << "Ret: " << ret << endl;
 if (ret > MAX) {
  cout << "User " << a << " already exists" << endl;
 } else {
  **Person *p = new Person(a);
  ppls[current] = p;**
  cout << "Added user: " << a << " at index: " << current << endl;
  current++;
 }
}

因此，现在的输出结果为：

ppls[0] is NULL
ppls[1] is NULL
ppls[2] is NULL
ppls[3] is NULL
ppls[4] is NULL
Added user: 21 at index: 0
21
Added user: 7 at index: 1
21 7
Added user: 42 at index: 2
21 7 42
User 21 already exists
Segmentation fault (core dumped)

什么是分段错误，如何修复它？谢谢。

- this is a long display name

你正在将堆栈变量的地址分配给指针数组...这总是一个坏主意。Person p(a); 然后 ppls[current] = &p;？我没有时间写一个适当的答案，但你在编程职业生涯中还有很长的路要走... - nonsensickle

谢谢。我知道。我仍然只是一个本科生。 - this is a long display name

抱歉如果我表现得不够尊重或冒犯了您。看到新人开始学习编程是好事，很抱歉我没有时间来提供适当的帮助。欢迎加入我们！ - nonsensickle

请阅读这篇博客文章，学习如何使用调试器。它们在未来将对您非常宝贵。 - nonsensickle

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- R Sahu · Accepted Answer

Person p(a);
ppls[current] = &p;

这是一个问题。您正在存储指向本地变量的指针。您的程序可能会出现未定义行为。

请使用

Person* p = new Person(a);
ppls[current] = p;

请确保在Ppl的析构函数中删除对象。

改进建议

不清楚您的程序目标是什么，但使用以下方法可以使您的生活更加简单：

std::vector<Person> ppls;

代替

Person *ppls[MAX];