Question

Why does the following code run?

#include <iostream>
class A {
    int num;
    public:
        void foo(){ num=5; std::cout<< "num="; std::cout<<num;}
};

int main() {
    A* a;
    a->foo();
    return 0;
}

The output is

num=5

I compile this using gcc and I get only the following compiler warning at line 10:

(warning: ‘a’ is used uninitialized in this function)

But as per my understanding, shouldn’t this code not run at all? And how come it’s assigning the value 5 to num when num doesn’t exist because no object of type A has been created yet?

Answer 1

You haven’t initialized *a.

Try this:

#include <iostream>

class A
{
    int num;
    public:
        void foo(){ std::cout<< "num="; num=5; std::cout<<num;}
};

int main()
{
    A* a = new A();
    a->foo();
    return 0;
}

---

Not initializing pointers (properly) can lead to undefined behavior. If you’re lucky, your pointer points to a location in the heap which is up for initialization*. (Assuming no exception is thrown when you do this.) If you’re unlucky, you’ll overwrite a portion of the memory being used for other purposes. If you’re really unlucky, this will go unnoticed.

This is not safe code; a “hacker” could probably exploit it.

*Of course, even when you access that location, there’s no guarantee it won’t be “initialized” later.

---

“Lucky” (actually, being “lucky” makes it more difficult to debug your program):

// uninitialized memory 0x00000042 to 0x0000004B
A* a;
// a = 0x00000042;
*a = "lalalalala";
// "Nothing" happens

---

“Unlucky” (makes it easier to debug your program, so I don’t consider it “unlucky”, really):

void* a;
// a = &main;
*a = "lalalalala";
// Not good. *Might* cause a crash.
// Perhaps someone can tell me exactly what'll happen?