CODEWITHSUNDEEP
c++operator-overloadingnew-operator
Bruno Gouveia
Bruno Gouveia

Reputation: 35

Overloading operator new / allocate more than object size

I'm trying to overloading the operator new of my class, I wanna allocate space for the objects' attributes and for a vector of chars that isn't defined in the class.

But when I instance an new object, the chars of previously object are changing. Am I doing something wrong?

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] != '\0' && len < (maxLen - 1))
        {
            data[len] = string[len];
            len++;
        }
        data[len] = '\0';
        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); 
    }
};

I use this in other class: 

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] != '\0')
    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("%s\n", data->buffer()); }

And my main function is: 

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;

}

When I execute, the strings are broken:

boisahzashdacaraverdeepretaeazuleamarelaecinzaevermelha

boisahzashdacaraverd

freestepehnoisquevoa

Upvotes: 2

Views: 304

Answers (1)

Mats Petersson
Mats Petersson

Reputation: 129454

This is not a great idea:

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

Not only does it fail if you ever add/remove some member functions [or make a virtual function], but it's also dependent on the size of int.

Use 

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

[Note newplacement of the end of reinterpret_cast<>()!!] The first version skips 12 * sizeof(*this) forwards, rather than 12 bytes. [Of course, on my machine, when I used the 4* multiplier of maxLen, it actually works - it only began failing when I removed that]

This can be simplified:

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

Is this supposed to be 4 times larger than maxlen:

(4 * maxLen)

to:

4 * (len / 4 + 1);

I will be back with edits to this answer once I know the answer to the comment about how it is being tested.

Upvotes: 1

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