As @JDługosz points out in the comments, Herb gives other advice in another (later?) talk, see roughly from here: https://youtu.be/xnqTKD8uD64?t=54m50s.
His advice boils down to only using value parameters for a function f
that takes so-called sink arguments, assuming you will move construct from these sink arguments.
This general approach only adds the overhead of a move constructor for both lvalue and rvalue arguments compared to an optimal implementation of f
tailored to lvalue and rvalue arguments respectively. To see why this is the case, suppose f
takes a value parameter, where T
is some copy and move constructible type:
void f(T x) {
T y{std::move(x)};
}
Calling f
with an lvalue argument will result in a copy constructor being called to construct x
, and a move constructor being called to construct y
. On the other hand, calling f
with an rvalue argument will cause a move constructor to be called to construct x
, and another move constructor to be called to construct y
.
In general, the optimal implementation of f
for lvalue arguments is as follows:
void f(const T& x) {
T y{x};
}
In this case, only one copy constructor is called to construct y
. The optimal implementation of f
for rvalue arguments is, again in general, as follows:
void f(T&& x) {
T y{std::move(x)};
}
In this case, only one move constructor is called to construct y
.
So a sensible compromise is to take a value parameter and have one extra move constructor call for either lvalue or rvalue arguments with respect to the optimal implementation, which is also the advice given in Herb's talk.
As @JDługosz pointed out in the comments, passing by value only makes sense for functions that will construct some object from the sink argument. When you have a function f
that copies its argument, the pass-by-value approach will have more overhead than a general pass-by-const-reference approach. The pass-by-value approach for a function f
that retains a copy of its parameter will have the form:
void f(T x) {
T y{...};
...
y = std::move(x);
}
In this case, there is a copy construction and a move assignment for an lvalue argument, and a move construction and move assignment for an rvalue argument. The most optimal case for an lvalue argument is:
void f(const T& x) {
T y{...};
...
y = x;
}
This boils down to an assignment only, which is potentially much cheaper than the copy constructor plus move assignment required for the pass-by-value approach. The reason for this is that the assignment might reuse existing allocated memory in y
, and therefore prevent (de)allocations, whereas the copy constructor will usually allocate memory.
For an rvalue argument the most optimal implementation for f
that retains a copy has the form:
void f(T&& x) {
T y{...};
...
y = std::move(x);
}
So, only a move assignment in this case. Passing an rvalue to the version of f
that takes a const reference only costs an assignment instead of a move assignment. So relatively speaking, the version of f
taking a const reference in this case as the general implementation is preferable.
So in general, for the most optimal implementation, you will need to overload or do some kind of perfect forwarding as shown in the talk. The drawback is a combinatorial explosion in the number of overloads required, depending on the number of parameters for f
in case you opt to overload on the value category of the argument. Perfect forwarding has the drawback that f
becomes a template function, which prevents making it virtual, and results in significantly more complex code if you want to get it 100% right (see the talk for the gory details).
std::string_view
there's never a good case for passing a reference to string. But given when you asked this question, that's probably not what you're looking for! – Dorathydorca