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C++ operator overloading
Posted: Fri Aug 26, 2011 5:41 pm
by adamemanuel
Hey geeks,
Have a n00bish question, I'm overloading the [] operator to implement an array like class.
It works fine, my problem arises when I've a pointer to the object.
MyClass c;
c[5] works fine
However
MyClass *c = new MyClass();
*c[5] will not work since the precedence of [] is higher than that of *
So I've to write it as
(*c)[5] which is not looking good for me
any solution to overcome this.
Re: C++ operator overloading
Posted: Fri Aug 26, 2011 10:10 pm
by schilds
If you like solutions which are more painful than the original problem:
Code: Select all
template< typename C, typename T >
class CollectionPointer
{
private:
C * collection;
public:
CollectionPointer(C * collection)
: collection(collection)
{
}
T& operator[](int i)
{
return (*collection)[i];
}
C& operator*()
{
return *collection;
}
C* operator->()
{
return collection;
}
void clean(){
delete collection;
}
};
#include <iostream>
#include <vector>
typedef CollectionPointer< std::vector<int>, int > MyVector;
int main()
{
MyVector numbers(new std::vector<int>());
numbers->push_back(0);
std::cout << numbers[0] << std::endl;
numbers.clean();
}
Tada!
Re: C++ operator overloading
Posted: Sat Aug 27, 2011 6:31 pm
by adamemanuel
That's what I was trying to achive:
Code: Select all
class MyArr{
public:
MyArr();
~MyArr();
uint8_t& operator[](const int);
private:
uint8_t *arr;
unsigned int size;
void init();
};
Code: Select all
MyArr::MyArr(){
size = 1000;
init();
}
MyArr::~MyArr(){
delete[] arr;
size = 0;
}
void MyArr::init(){
arr = new uint8_t[size];
}
uint8_t& MyArr::operator[](const int i){
if(){
// some checkings
}
return arr[i];
}
Code: Select all
int main(){
MyArr m;
m[2] = 5; // <--- works fine
MyArr *k = new MyArr();
(*k)[2] = 3; // <--- I want to be able to just use *k[2] = 3 or k[2] = 3 instead !!!!
uint8_t y = (*k)[2];
std::cout << (int)y << std::endl;
return 0;
}