c++ variable scope

[t048]

If you declare a variable in a function

void myfunc (){
 * * *int a = 33;
}

that variable (in this case [tt]a[/tt]) only exists throughout the duration of the function and is then wiped off the stack.

I assume it works the same way with objects

void myfunc(){
 * * *MyObject obj;
}

[tt]obj[/tt] only exists on the stack throughout the function.

My question is what happens here:

void myfunc(){
 * * *MyObject *obj = new MyObject;
}

The [tt]new[/tt] operator creates the object on the heap, so the object remains after the function is over, right? *But wouldn't the pointer to that object be removed along with all other local variables when the function ends? *So this object would be a memory leak, right? *

[TaeShadow]

The [tt]new[/tt] operator creates the object on the heap, so the object remains after the function is over, right? But wouldn't the pointer to that object be removed along with all other local variables when the function ends? So this object would be a memory leak, right?

That is correct.

[t048]

Ok, so this should work:

void myfunc(){

 * * *MyObject *obj = getObj ();
}

MyObject *getObj (){
 * * *
 * * *MyObject *anotherObj = new MyObject;
 * * *return anotherObj;
}

The object is created in the [tt]getObj[/tt] function and continues to exist when the function is over and the pointer is returned to the [tt]myfunc[/tt] function. *This wouldn't cause any errors, right?

[TaeShadow]

The object is created in the [tt]getObj[/tt] function and continues to exist when the function is over and the pointer is returned to the [tt]myfunc[/tt] function. This wouldn't cause any errors, right?

Well, when myfunc() returns, all pointers to the object will have expired, and there will be a leak, but other than that it looks good.

[t048]

Well, when myfunc() returns, all pointers to the object will have expired, and there will be a leak, but other than that it looks good.

Oh, yes, of course. I was referring to the getobj function.