Why is the scope of object created using new limited? [duplicate] - c++

Question

This question already has answers here:
Do Pointer Parameters Need To Be Passed By Reference
(3 answers)
Closed 3 years ago.
Object made using new operator does not seem to be available outside the scope! Isn't that the whole point of the new operator?
https://ideone.com/DDvo9y - Please check this link to see the result.
#include<iostream>
class myClass
{
private:
int val;
public:
myClass () = delete;
myClass (int val):val{val}{}
int get () const
{
return val;
}
};
bool ifEqualMake (int a, int b, myClass * obj)
{
if (a == b) obj = new myClass (a);
else{
std::cout << "Difference exists: " << a - b << '\n';
obj = new myClass (a + b);
}
std::cout << " Object made with value :" << obj->get () << '\n';
return (a == b);
}
int main ()
{
myClass *obj1 = nullptr;
myClass *obj2 = nullptr;
myClass *obj3 = nullptr;
ifEqualMake (3, 3, obj1);
ifEqualMake (4, 3, obj2);
ifEqualMake (4, 4, obj3);
if(obj1) std::cout << "obj 1 made in heap: " << obj1->get () << '\n';
if(obj2) std::cout << "obj 2 made in heap: " << obj2->get()<<'\n';
if(obj3) std::cout << "obj 3 made in heap: " << obj3->get () << '\n';
delete obj1;
delete obj2;
delete obj3;
return 0;
}

Answer 1

It isn't.
You're confusing the dynamically-allocated object that you created with new, and the pointer that points to it.
The pointer's scope is limited just like any other automatic-storage-duration object.
It looks like you meant to use it as an "out" argument to the function ifEqualMake, perhaps by taking a reference to it rather than a copy. Then alterations to it, such as pointing it to a new object, will be mirrored in the calling scope.

Answer 2

The parameter obj is passed by value, that means it's just a copy of the argument, and any modification on itself inside the function (like obj = new myClass (a);) has nothing to do with the original pointer. In the meanwhile, the object constructed inside the function won't be destroyed.
You might change it to pass-by-reference.
bool
ifEqualMake (int a, int b, myClass *& obj)
// ^
{
...
}

Answer 3

Consider the following function:
void foo(int i) {
i = 4;
}
void bar() {
int j = 0;
foo(j);
std::cout << j << '\n';
}
You would expect bar to print 0, not 4, because foo is assigning to a local variable.
This behaviour does not change with pointers. The following code behaves the same way:
void foo(int* i) {
int temp = 0;
i = &temp;
}
void bar() {
int* j = nullptr;
foo(j);
std::cout << j << '\n';
}
The simplest fix to the code you present is to take the pointer by reference:
void foo(int*& i) {
int temp = 0;
i = &temp;
}
void bar() {
int* j = nullptr;
foo(j);
// j now points to a destroyed object
std::cout << j << '\n';
}