How to transfer object to function as it's class'es member - c++

Question

For example: I have class "ListNode" which is linked list member, and i overloaded a [] operator for List class itself, but every time i need to get data from a node, i need to write list[0]->getData() or if I want to set new data, I need to write list[0]->setData(*somedata*).
But I want to make it be able to work like for example like list[0] = 5, so first Node's data is setted to 6. Same with functions. What should I do if I have Node which has int data, and I need to transfer it's data to a function, but without writing Node->getData() all the time?

Answer 1

You can make your operator[] return a reference to the element you want to access.
The example below is a simplified version of what would be a proper solution, including a const version for the operator[].
[Demo]
#include <iostream> // cout
#include <stdexcept> // range_error
struct Node
{
int value{};
Node* next{};
int& operator[](size_t pos)
{
if (pos == 0) { return this->value; }
else if (next == nullptr) { throw std::range_error{"out of bounds"}; }
else { return (*next)[pos - 1]; }
}
};
int main()
{
Node n5{5, nullptr};
Node n4{4, &n5};
Node n3{3, &n4};
Node n2{2, &n3};
Node list{1, &n2};
try
{
std::cout << "list[3]: " << list[3] << "\n";
list[3] = 25;
std::cout << "list[3]: " << list[3] << "\n";
std::cout << "list[13]: " << list[13] << "\n";
}
catch (const std::exception& exc)
{
std::cout << "Error: " << exc.what() << ".\n";
}
}