内存分配 - 正确使用删除

Question

由于delete [] meanings;、delete [] meanings;、delete [] temp_meaning;，我的所有程序都崩溃了，当我删除这 3 行时它工作正常，所以可能我错误地使用了 delete ... 有人可以在这里启发我吗？

#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(const 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 * [meanings_ctr]; // Allocate Space for meanings
}

Expression::~Expression() {

    while(meanings_ctr-->0){
    delete meanings[meanings_ctr];
    }
    delete [] meanings; // Deleting the memory we allocated
    delete [] word_with_several_meanings; // Deleting the memory we allocated
}

void Expression::word(const 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[ meanings_ctr ] = new char [strlen(p2c) + 1];
    //strcpy(meanings[ meanings_ctr++ ] , p2c);
    // temp 
    if (meanings_ctr < 1){
    meanings[ meanings_ctr ] = new char [strlen(p2c) + 1];
    strcpy(meanings[ meanings_ctr++ ] , p2c);
    }
    else {
int temp_ctr;
    char **temp_meaning;
    temp_meaning = new char * [meanings_ctr-1];
    for(temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
        temp_meaning[temp_ctr] = new char [strlen(meanings[ temp_ctr ]) + 1];
            strcpy(temp_meaning[temp_ctr], meanings[ temp_ctr ]);
    }
    for (temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
            delete meanings[temp_ctr];

    }
    delete [] meanings;

    meanings = new char * [meanings_ctr];
    for(temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
        meanings[ temp_ctr ] = new char [strlen(temp_meaning[temp_ctr]) + 1];
            strcpy(meanings[ temp_ctr ], temp_meaning[temp_ctr]);
    }
    meanings[ meanings_ctr ] = new char [strlen(p2c) + 1];
    strcpy(meanings[ meanings_ctr ] , p2c);
            for (temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
            delete temp_meaning[temp_ctr];
    }
    delete [] temp_meaning;
            meanings_ctr++;
    }


}

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 very long test");
    char str[] = "Meaning_    ";
    for(int kx =0; kx<31; kx++){
            str[8] = ('A'+kx);
            expr3.add_meaning(str);
    }

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

Answer 1

该程序显示出来自以下语句的内存损坏迹象：

    meanings_ctr = mctr;    // Setting the counter to 0
    meanings = new char * [meanings_ctr]; // Allocate Space for meanings

和

    meanings = new char * [meanings_ctr];

由于 add_meaning() 包含以下代码：

if (meanings_ctr < 1){
   meanings[ meanings_ctr ] = new char [strlen(p2c) + 1];

您实际上是在含义[0] 上写的，而您为它分配了 0 个字节。由于 C 中的索引从 0 开始，对于在 max_index 具有最高索引的数组，您需要分配 max_index+1 元素。对于 max_index = 0 的数组，您需要分配 1 个元素。

换句话说，您需要分配meanings = new char * [meanings_ctr + 1] 而不是new char * [meanings_ctr]，以及temp_meaning = new char * [meanings_ctr] 而不是new char * [meanings_ctr - 1]。

对于delete和delete[]的使用，一般规则是new分配的应该用delete释放，new[]分配的应该用delete[]销毁。上面有一个线程：delete vs delete[] operators in C++。在how does delete know it is an array的答案中可以找到一些好的背景。

以下是如何在不使用任何昂贵的工具或难以学习的工具的情况下调试程序。

如果您将调试打印添加到构造函数和析构函数中，如下所示：

Expression::Expression(int mctr ) {
    meanings_ctr = mctr;    // Setting the counter to 0
    meanings = new char * [meanings_ctr]; // Allocate Space for meanings
    cout << "[debug] allocated " << sizeof(char*)*meanings_ctr << " bytes @" << 
             hex << meanings << dec << endl;
}

还有

Expression::~Expression() {
    while(meanings_ctr-- > 0){
//       if(meanings[meanings_ctr]) delete [] (meanings[meanings_ctr]);
    }
    cout << "[debug] to deallocate @" << hex << meanings << dec << endl;
//   delete [] meanings; // Deleting the memory we allocated
//    delete [] word_with_several_meanings; // Deleting the memory we allocated
}

同样在 add_meaning() 中，你得到 ​​p>

[debug] allocated 0 bytes @0x804c008
[debug] to deallocate @0x804c008
[debug] allocated 4 bytes @0x804c078
 ...
[debug] allocated 120 bytes @0x804fa78
[debug] to deallocate @0x804fa78
[debug] to deallocate @0x804c260
[debug] to deallocate @0x804c150

这里看起来令人担忧的是allocated 0 bytes。由于 add_meanings() 中的代码包含：

if (meanings_ctr < 1){
meanings[ meanings_ctr ] = new char [strlen(p2c) + 1];

它使用未分配的内存@meanings[0] 并导致损坏。

这里是所有累积变化的参考：

25c25
<     meanings = new char * [meanings_ctr]; // Allocate Space for meanings
---
>     meanings = new char * [meanings_ctr + 1]; // Allocate Space for meanings
30,31c30,31
<     while(meanings_ctr-->0){
<     delete meanings[meanings_ctr];
---
>     while(meanings_ctr-- > 0){
>         delete [] meanings[meanings_ctr];
58c58
<     temp_meaning = new char * [meanings_ctr-1];
---
>     temp_meaning = new char * [meanings_ctr ];
64c64
<             delete meanings[temp_ctr];
---
>          delete [] meanings[temp_ctr];
69c69
<     meanings = new char * [meanings_ctr];
---
>     meanings = new char * [meanings_ctr + 1];
76,77c76,77
<             for (temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
<             delete temp_meaning[temp_ctr];
---
>     for (temp_ctr =0; temp_ctr<meanings_ctr;temp_ctr++){
>            delete [] temp_meaning[temp_ctr];

Answer 2

你必须来自 C 世界，你不会在 C++ 程序中找到任何 char* 来存储单词...你应该看看 std::string，它会启发你的一天，我可以向你保证.

顺便说一句，在 C++ 中，我们倾向于摆脱所有那些 delete 和 delete []，您也许应该选择一本好的 C++ 书籍并学习正确的 C++，而不是 30 年前使用的带有类的 C。