Hard Copy
出现原因:多个对象同时指向一块内存空间。
解决方案:所有的对象都拥有自己的内存数据。
11.2中的例子中,错误的罪魁祸首就是C++默认提供的Copy Constructor,它把指针直接赋值了。
- Provide a user-defined copy constructor. ```cpp MyString::MyString(const MyString & ms) { this->buf_len = 0; this->characters = NULL; create(ms.buf_len, ms.characters); }
- `create()` release the current memory and **allocate a new one.**
- **会显示放当然内存,然后重新申请一个新的内存空间。**
- `**t**his->characters` will not point to `ms.characters`.
- 保证了当前对象的指针不跟参数的指针相同
- **It's a hard copy!**
第二个错误产生的原因:赋值运算符出现了问题<br />默认的运算符重载做了两个变量的复制,我们可以自定义运算符重载函数,
- Provide a user-defined copy assignment
- 使用create去申请内存,使得每个对象都有自己的内存,不在共享
```cpp
#pragma once
#include <iostream>
#include <cstring>
class MyString
{
private:
int buf_len;
char *characters;
public:
MyString(int buf_len = 64, const char *data = NULL)
{
std::cout << "Constructor(int, char*)" << std::endl;
this->buf_len = 0;
this->characters = NULL;
create(buf_len, data);
}
MyString(const MyString &ms)
{
std::cout << "Constructor(MyString&)" << std::endl;
this->buf_len = 0;
this->characters = NULL;
create(ms.buf_len, ms.characters);
}
~MyString()
{
release();
}
MyString &operator=(const MyString &ms)
{
create(ms.buf_len, ms.characters);
return *this;
}
bool create(int buf_len, const char *data)
{
release();
this->buf_len = buf_len;
if (this->buf_len != 0)
{
this->characters = new char[this->buf_len]{};
}
if (data)
strncpy(this->characters, data, this->buf_len);
return true;
}
bool release()
{
this->buf_len = 0;
if (this->characters != NULL)
{
delete[] this->characters;
this->characters = NULL;
}
return 0;
}
friend std::ostream &operator<<(std::ostream &os, const MyString &ms)
{
os << "buf_len = " << ms.buf_len;
os << ", characters = " << static_cast<void *>(ms.characters);
os << " [" << ms.characters << "]";
return os;
}
};