[c++] C++ Returning reference to local variable

Is the following code (func1()) correct if it has to return i? I remember reading somewhere that there is a problem when returning reference to a local variable. How is it different from func2()?

int& func1()
{
    int i;
    i = 1;
    return i;
}

int* func2()
{
    int* p;
    p = new int;
    *p = 1;
    return p;
}

A local variable is memory on the stack, that memory is not automatically invalidated when you go out of scope. From a Function deeper nested (higher on the stack in memory), its perfectly safe to access this memory.

Once the Function returns and ends though, things get dangerous. Usually the memory is not deleted or overwritten when you return, meaning the memory at that adresss is still containing your data - the pointer seems valid.

Until another function builds up the stack and overwrites it. This is why this can work for a while - and then suddenly cease to function after one particularly deeply nested set of functions, or a function with really huge sized or many local objects, reaches that stack-memory again.

It even can happen that you reach the same program part again, and overwrite your old local function variable with the new function variable. All this is very dangerous and should be heavily discouraged. Do not use pointers to local objects!

A good thing to remember are these simple rules, and they apply to both parameters and return types...

• Value - makes a copy of the item in question.
• Pointer - refers to the address of the item in question.
• Reference - is literally the item in question.

There is a time and place for each, so make sure you get to know them. Local variables, as you've shown here, are just that, limited to the time they are locally alive in the function scope. In your example having a return type of int* and returning &i would have been equally incorrect. You would be better off in that case doing this...

void func1(int& oValue)
{
    oValue = 1;
}

Doing so would directly change the value of your passed in parameter. Whereas this code...

void func1(int oValue)
{
    oValue = 1;
}

would not. It would just change the value of oValue local to the function call. The reason for this is because you'd actually be changing just a "local" copy of oValue, and not oValue itself.