重载操作符new / allocate大于对象大小

Overloading operator new / allocate more than object size

本文关键字:对象 大于 allocate 操作符 new 重载      更新时间:2023-10-16

我试图重载我的类的operator new,我想为对象的属性和在类中未定义的字符向量分配空间。

但是当我实例化一个新对象时,前一个对象的字符正在改变。我做错了什么吗?

class StringData
{       
public:
    friend class String;
    int refCount;
    int len;
    int maxLen;
    ~StringData()
    {
    }
    StringData()
    {
        maxLen = 1000;
        len = 0;
        refCount = 1;
    }
    StringData(const char * string)
    {
        maxLen = 1000;
        char * data = buffer();
        len = 0;
        while (string[len] != '' && len < (maxLen - 1))
        {
            data[len] = string[len];
            len++;
        }
        data[len] = '';
        refCount = 1;
    }
    char* buffer()
    {
        return reinterpret_cast<char*>(this + 12);
    }
public:
    void* operator new(size_t size, int maxLen)
    {
        return ::operator new(size + round4(maxLen + 1));
    }
    static int round4(int len)
    {
        return ((((int) (len / 4)) + 1) * 4);
    }
    void operator delete(void* obj)
    {
        ::operator delete(obj); 
    }
    void operator delete(void* obj, int size) throw ()
    {
        ::operator delete(obj); 
    }
};

我在其他类中使用:

class String{
public:
    StringData * data;
    String(StringData * data){
        this->data = data;
    }
public:
    String(const char*);
    String(const String&);
    ~String();
    String& operator =(const String);
    String& operator =(const char *);
    int length() const;
    bool operator ==(const String&) const;
    int compare(const String&) const;
    bool operator ==(const char*) const;
    int compare(const char*) const;
    String& operator +(const String&) const;
    String operator +(const char*) const;
    String operator +=(const String&);
    String operator +=(const char*);
    String toLower();
    String toUpper();
    char operator [ ](int) const;
    char& operator [ ](int);
    void print() const;

};

String::String(const char * string){
int stringLen = 0;
while (string[stringLen] != '')
    stringLen++;
data = new (stringLen + 1) StringData(string);

}

String::String(const String& string){
data = string.data;
string.data->refCount++;

}

String::~String(){
this->data->refCount--;
if (this->data->refCount == 0)
    delete data;

}

String& String::operator=(const String string){
data->refCount--;
if (this->data->refCount == 0)
    delete data;
data = string.data;
string.data->refCount++;
return *this;

}

void String::print() const{ printf("%sn", data->buffer()); }

我的主要功能是:

int main(){
String *a = new String("boisahzashdacaraverdeepretaeazuleamarelaecinzaevermelha");
a->print();
String * s = new String("freestepehnoisquevoaashashashhasshashhasssasassadasd");
String * b = new String("kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk");
a->print();
s->print();
return 0;

}

当我执行时,字符串被打破了:

boisahzashdacaraverdeepretaeazuleamarelaecinzaevermelha

boisahzashdacaraverd

freestepehnoisquevoa

这不是一个好主意:

return reinterpret_cast<char*>(this + 12);

不仅在添加/删除一些成员函数[或创建虚函数]时失败,而且还依赖于int的大小。

使用

return reinterpret_cast<char*>(this) + sizeof(*this);

[注:reinterpret_cast<>()结尾的newplacement !!]第一个版本向前跳过12 * sizeof(*this)字节,而不是跳过12字节。[当然,在我的机器上,当我使用maxLen的4*乘数时,它实际上是有效的-只有当我删除它时才开始失败]

可以简化为:

((((int) (len / 4)) + 1) * 4);

这个应该是maxlen:

的4倍大吗?
(4 * maxLen)

:

4 * (len / 4 + 1);
一旦我知道关于它是如何被测试的评论的答案,我会回来编辑这个答案。