Question

I created a class A and wrote the following function foo()

class A
{
public:
int a;
};

A * foo()
{
A a1;
return &a1;
}

int main()
{
A * a2;
a2 = foo();
return 0;
}

The compiler gave me a warning as a1 is a local variable and I am returning its address from the stack (so its value can change unpredictably).

Now I changed foo() to the following

A * foo()
{
A a1;
A *a3;
a3 = &a1;
return a3;
}

Now the compiler does not give any warning. Is this because a3 is created on the heap? If so are pointers always created on heap like this. I thought heap is utilized only through new/malloc.

Answer 1

Now the compiler does not give any warning.

The compiler isn’t giving any warning because you’ve added sufficient complexity to fool the analysis that it does of your code.

You are still returning a pointer to a local variable, and you cannot use that pointer after the function has returned.