字典的動態內存分配

Question

嗨，我需要建立一個像字典一樣，根據我的代碼每個單詞可以有100個含義，但也許它只有5個含義，那麼我將分配95額外的空間沒有或可能它有超過100個含義，那麼程序會崩潰，我知道矢量類很容易，可以很好地使用，但任務幾乎建立我自己的矢量類，以瞭解它是如何工作的。因此，**意義和其他一些東西保持不變，這裏是我的代碼，也知道我造成內存泄漏，我該如何正確刪除？ ：

#include <iostream> 
#include <string> 
#include <cstring> 
using namespace std; 

class Expression { 

    char *word_with_several_meanings; // like "bank", "class" 
    char **meanings; // a pointer to a pointer stores all meanings 
    int meanings_ctr; // meanings counter 

    //-----------FUNCTIONS------------------------------------------------ 
public: 
    void word(char* = NULL); 
    void add_meaning(char* = NULL); 
    char* get_word(); 
    int get_total_number_of_meanings(); 
    char* get_meaning(int meanx = 0); 
    Expression(int mctr = 0); // CTOR 
    ~Expression(); // DTOR 
}; 

    Expression::Expression(int mctr) { 
    meanings_ctr = mctr;   // Setting the counter to 0 
    meanings = new char * [100]; // Allocate Space for 100 meanings 
} 

Expression::~Expression() { 
delete [] meanings; // Deleting the memory we allocated 
delete [] word_with_several_meanings; // Deleting the memory we allocated 
} 

void Expression::word(char *p2c) 
{ 

    word_with_several_meanings = new char[strlen(p2c)+1]; 
// copy the string, DEEP copy 
    strcpy(word_with_several_meanings, p2c); 
} 

void Expression::add_meaning(char *p2c) 
{ 

    //meanings = new char * [meanings_ctr+1]; 
    meanings[meanings_ctr] = new char[strlen(p2c)+1]; 
    strcpy(meanings[meanings_ctr++],p2c); 


} 

char * Expression::get_meaning(int meanx) 
{ 

    return *(meanings+meanx); 

} 

char * Expression::get_word() 
{ 

    return word_with_several_meanings; 

} 

int Expression::get_total_number_of_meanings() 
{ 

    return meanings_ctr; 

} 

int main(void) { 
    int i; 
    Expression expr; 
    expr.word("bank "); 
    expr.add_meaning("a place to get money from"); 
    expr.add_meaning("b place to sit"); 
    expr.add_meaning("4 letter word"); 
    expr.add_meaning("Test meaning"); 
    cout << expr.get_word() << endl; 

    for(int i = 0; i<expr.get_total_number_of_meanings(); i++) 
      cout << " " << expr.get_meaning(i) << endl; 
    Expression expr2; 
    expr2.word("class"); 
    expr2.add_meaning("a school class"); 
    expr2.add_meaning("a classification for a hotel"); 
    expr2.add_meaning("Starts with C"); 
    cout << expr2.get_word() << endl; 
    for(i = 0; i<expr2.get_total_number_of_meanings(); i++) 
      cout << " " << expr2.get_meaning(i) << endl; 

     Expression expr3; 
    expr3.word("A long test ... "); 
    char str[] = "Meaning_  "; 
    for (int kx=0;kx<26;kx++) 
    { 
      str[8] = (char) ('A'+kx); 
      expr3.add_meaning(str); 
    } 

cout << expr3.get_word() << endl; 
for(i = 0; i < expr3.get_total_number_of_meanings(); i++) 
    cout << " " << expr3.get_meaning(i) << endl; 

    return 0; 
} 

Answer 1

當您分配與new則您與一個循環，例如分配它的多維陣列

char **x = new char*[size] 
for (int i = 0; i < N; i++) { 
    x[i] = new int[size]; 
} 

所以你也必須刪除它以這種方式：

for (int i = 0; i < N; i++) { 
    delete[] x[i]; 
} 
delete[] x; 

因此，當你遇到你的數組的任意大小你必須給他們一些地方保存析構函數中使用它們。

Answer 2

delete [] meanings; // Deleting the memory we allocated 

不會擺脫你分配的內存，只有指針本身。

要釋放實際內存，您需要遍歷您的meanings陣列和delete []中的每個元素。

喜歡的東西：

for (int i = 0; i < meanings_ctr; ++i) 
{ 
    delete [] meanings[meanings_ctr]; 
    meanings[meanings_ctr] = NULL; 
} 
delete [] meanings; 

-

有關，如果你得到超過100個含義做什麼的問題（或者一般而言，當您的收藏已滿），標準的方法是分配一個尺寸加倍的新數組（因爲它是動態的，你可以這樣做），將現有的集合複製到那個數組中，然後處理現有的數組。

Answer 3

我會使用一個簡單的鏈表（這是簡化的，不完整的，未經檢驗，也應該有適當的getter/setter方法和材料）：

class Meaning { 
    char text[20]; 
    Meaning *next; 

    Meaning(const char *text) : next(0) { 
     strcpy(this->text, text); 
    } 
} 

class Word { 
    char text[20]; 
    Meaning *first; 
    Meaning *last; 

    Word(const char *text) : first(0), last(0) { 
     strcpy(this->text, text); 
    } 

    ~Word() { 
     Meaning *m = first, *n; 
     while(m) { 
      n = m->next; 
      delete m; 
      m = n; 
     } 
    } 

    void AddMeaning(const char *text) { 
     if (last) { 
      last = last->next = new Meaning(text); 
     } 
     else { 
      first = last = new Meaning(text); 
     } 
    } 

    void print() { 
     printf("%s:\n\t", text); 
     Meaning *m = first; 
     while (m) { 
      printf("%s, ", m->text); 
      m = m->next; 
     } 
    } 
}