Pass by constant reference

Pass by constant reference ensures that the argument won’t be modified BY THE FUNCTION. So if in the function you push the argument onto a vector then return from the function, you then have the argument stored in the vector and you can do whatever you want with it, right?

Yes, but you will be doing whatever you want with a copy of the object. Using a vector for this is quite unnecessary.

Okay so in this example, where I'm simulating STL vectors with
limit = a pointer to the memory location AFTER the last available one
avail = a pointer to the next available memory location

Code:

---

template<typename T>
void Vec<T>::push_back(const T & val)
{
if (avail == limit)
grow();
*avail = val;
++avail;
}

---

would the location pointed to by avail store the argument or a copy of it?

You would store a copy of it since you dereference the pointer and you assign the object itself and not its address.

Right... thanks!