Error: cannot convert from 'long' to 'int &' - c++

Question

my function get_num returns a variable of long int.
I want to write the next operator:
long int& operator [] (long int i) {
long int a = get_num(i);
int& b = a;
return b;
}
but I got the next error:
error C2440: 'initializing' : cannot convert from 'long' to 'int &'
error C2440: 'return' : cannot convert from 'int' to 'long &'
how can I fix it without changing the function of get_num?
any help appreciated!
I have a class:
class B {
B* next;
long int nom;
long int denom;
public:
long int get_nom() {return nom; }
long int get_denom() {return denom; }
};
class List {
B* head;
public:
long int& operator [] (long int desired_denom) {
// here I search the node that containts the denom that is equal to desired_denom
// and insert it to tmp (it's a pointer to B).
long int a1 = tmp->get_nom()
long int& a2 = a1;
return a2;
}
};
now in the main, I want to do:
int main() {
A a; // assume that it creates the list of B and put values in each node (each B)
// here I want to do:
a[2] = 3; // it should search the node that his denom is equal to 2, and puts 3 instead of his nom
return 0;
}
for example:
if my list is:
(nom=5, denom=6)->(nom=1,denom=8)->(nom=4, denom=2)->NULL
the line a[2]=3 searches the node that his denom is 2 (it's the third node) and set his nom to 3.
so after this line, my list will be:
(nom=5, denom=6)->(nom=1,denom=8)->(nom=3, denom=2)->NULL

Answer 1

Currently in your code there isn't a way to modify the nodes the way you want to. get_nom will have to be changed to return a reference before it can behave the way you would like. Right now you can only make copy of the values in each node of the linked list.
Assuming tmp is set correctly you can return the reference returned by get_nom
long int& operator [] (long int i) {
//Get tmp here probably by list traversal
return tmp->get_nom(i);
}
And then the implementation of get_nom. It is important to note that nom must be a value with a lifespan long enough to be of use as explained below. Returning nom which is private to the class and therefor will not fall out of scope when the function returns should work.
long int& get_nom() {return nom; }
get_nom must return long int& for this to work. There is an important difference between returning long int and long int&. When the return type is long int a copy of the value you are returning is made and passed to the caller. When you return long int& or in other words return a long int by reference, the reference points to the location in memory where the value is stored. This reference allows you to change the value of a long int while it is still stored in the middle of an array or whatever data type you are using.
Another important side effect of this is that if you return a reference to a local variable, when the variable falls out of scope, the value will no longer exist in memory. I haven't actually tried compiling and running this so I'm not sure if the compiler will get mad or what exactly the result will be, but I know it won't be good:
long int& operator [] (long int i) {
long int a = i + 5; //arbitrary change
return a;
}
In this case you are returning a reference to the location of a, but because a falls out of scope after the function finishes executing, the reference no longer points to anything at all. The variable you are returning a reference to must still exist after the function exits.

Answer 2

The compile error is for this line
int& b = a;
^^^^
It should be
long int& b = a;
^^^^^^^^
And, as friends have said in the comments, returning a reference to a local variable leads to a undefined-behavior because that referenced variable will be destroyed after exiting the function.
Blindly asnwering, you might need return a reference to the get_num, if get_num is returning a reference correctly.
return get_num(i);