In traditional C++, passing by value into functions and methods is slow for large objects, and is generally frowned upon. Instead, C++ programmers tend to pass references around, which is faster, but which introduces all sorts of complicated questions around ownership and especially around memory management (in the event that the object is heap-allocated)
Now, in C++11, we have Rvalue references and move constructors, which mean that it's possible to implement a large object (like an std::vector) that's cheap to pass by value into and out of a function.
So, does this mean that the default should be to pass by value for instances of types such as std::vector and std::string? What about for custom objects? What's the new best practice?
It's a reasonable default if you need to make a copy inside the body. This is what Dave Abrahams is advocating:
Guideline: Don’t copy your function arguments. Instead, pass them by value and let the compiler do the copying.
In code this means don't do this:
void foo(T const& t)
{
auto copy = t;
// ...
}
but do this:
void foo(T t)
{
// ...
}
which has the advantage that the caller can use foo like so:
T lval;
foo(lval); // copy from lvalue
foo(T {}); // (potential) move from prvalue
foo(std::move(lval)); // (potential) move from xvalue
and only minimal work is done. You'd need two overloads to do the same with references, void foo(T const&); and void foo(T&&);.
With that in mind, I now wrote my valued constructors as such:
class T {
U u;
V v;
public:
T(U u, V v)
: u(std::move(u))
, v(std::move(v))
{}
};
Otherwise, passing by reference to const still is reasonable.
In almost all cases, your semantics should be either:
bar(foo f); // want to obtain a copy of f
bar(const foo& f); // want to read f
bar(foo& f); // want to modify f
All other signatures should be used only sparingly, and with good justification. The compiler will now pretty much always work these out in the most efficient way. You can just get on with writing your code!
Pass parameters by value if inside the function body you need a copy of the object or only need to move the object. Pass by const& if you only need non-mutating access to the object.
Object copy example:
void copy_antipattern(T const& t) { // (Don't do this.)
auto copy = t;
t.some_mutating_function();
}
void copy_pattern(T t) { // (Do this instead.)
t.some_mutating_function();
}
Object move example:
std::vector<T> v;
void move_antipattern(T const& t) {
v.push_back(t);
}
void move_pattern(T t) {
v.push_back(std::move(t));
}
Non-mutating access example:
void read_pattern(T const& t) {
t.some_const_function();
}
For rationale, see these blog posts by Dave Abrahams and Xiang Fan.
The signature of a function should reflect it's intended use. Readability is important, also for the optimizer.
This is the best precondition for an optimizer to create fastest code - in theory at least and if not in reality then in a few years reality.
Performance considerations are very often overrated in the context of parameter passing. Perfect forwarding is an example. Functions like emplace_back are mostly very short and inlined anyway.
Related
Recently I find myself often in the situation of having a single function that takes some object as a parameter. The function will have to copy that object.
However the parameter for that function may also quite frequently be a temporary and thus I want to also provide an overload of that function that takes an rvalue reference instead a const reference.
Both overloads tend to only differ in that they have different types of references as argument types. Other than that they are functionally equivalent.
For instance consider this toy example:
void foo(const MyObject &obj) {
globalVec.push_back(obj); // Makes copy
}
void foo(MyObject &&obj) {
globalVec.push_back(std::move(obj)); // Moves
}
Now I was wondering whether there is a way to avoid this code-duplication by e.g. implementing one function in terms of the other.
For instance I was thinking of implementing the copy-version in terms of the move-one like this:
void foo(const MyObject &obj) {
MyObj copy = obj;
foo(std::move(copy));
}
void foo(MyObject &&obj) {
globalVec.push_back(std::move(obj)); // Moves
}
However this still does not seem ideal since now there is a copy AND a move operation happening when calling the const ref overload instead of a single copy operation that was required before.
Furthermore, if the object does not provide a move-constructor, then this would effectively copy the object twice (afaik) which defeats the whole purpose of providing these overloads in the first place (avoiding copies where possible).
I'm sure one could hack something together using macros and the preprocessor but I would very much like to avoid involving the preprocessor in this (for readability purposes).
Therefore my question reads: Is there a possibility to achieve what I want (effectively only implementing the functionality once and then implement the second overload in terms of the first one)?
If possible I would like to avoid using templates instead.
My opinion is that understanding (truly) how std::move and std::forward work, together with what their similarities and their differences are is the key point to solve your doubts, so I suggest that you read my answer to What's the difference between std::move and std::forward, where I give a very good explanation of the two.
In
void foo(MyObject &&obj) {
globalVec.push_back(obj); // Moves (no, it doesn't!)
}
there's no move. obj is the name of a variable, and the overload of push_back which will be called is not the one which will steal reasources out of its argument.
You would have to write
void foo(MyObject&& obj) {
globalVec.push_back(std::move(obj)); // Moves
}
if you want to make the move possible, because std::move(obj) says look, I know this obj here is a local variable, but I guarantee you that I don't need it later, so you can treat it as a temporary: steal its guts if you need.
As regards the code duplication you see in
void foo(const MyObject &obj) {
globalVec.push_back(obj); // Makes copy
}
void foo(MyObject&& /*rvalue reference -> std::move it */ obj) {
globalVec.push_back(std::move(obj)); // Moves (corrected)
}
what allows you to avoid it is std::forward, which you would use like this:
template<typename T>
void foo(T&& /* universal/forwarding reference -> std::forward it */ obj) {
globalVec.push_back(std::forward<T>(obj)); // moves conditionally
}
As regards the error messages of templates, be aware that there are ways to make things easier. for instance, you could use static_asserts at the beginning of the function to enfornce that T is a specific type. That would certainly make the errors more understandable. For instance:
#include <type_traits>
#include <vector>
std::vector<int> globalVec{1,2,3};
template<typename T>
void foo(T&& obj) {
static_assert(std::is_same_v<int, std::decay_t<T>>,
"\n\n*****\nNot an int, aaarg\n*****\n\n");
globalVec.push_back(std::forward<T>(obj));
}
int main() {
int x;
foo(x);
foo(3);
foo('c'); // errors at compile time with nice message
}
Then there's SFINAE, which is harder and I guess beyond the scope of this question and answer.
My suggestion
Don't be scared of templates and SFINAE! They do pay off :)
There's a beautiful library that leverages template metaprogramming and SFINAE heavily and successfully, but this is really off-topic :D
A simple solution is:
void foo(MyObject obj) {
globalVec.push_back(std::move(obj));
}
If caller passes an lvalue, then there is a copy (into the parameter) and a move (into the vector). If caller passes an rvalue, then there are two moves (one into parameter and another into vector). This can potentially be slightly less optimal compared to the two overloads because of the extra move (slightly compensated by the lack of indirection) but in cases where moves are cheap, this is often a decent compromise.
Another solution for templates is std::forward explored in depth in Enlico's answer.
If you cannot have a template and the potential cost of a move is too expensive, then you just have to be satisfied with some extra boilerplate of having two overloads.
The following is an example of my typical code. A have a lot of objects that look like this:
struct Config
{
Config();
Config(const std::string& cType, const std::string& nType); //additional variables omitted
Config(Config&&) = default;
Config& operator=(Config&&) = default;
bool operator==(const Config& c) const;
bool operator!=(const Config& c) const;
void doSomething(const std::string& str);
bool doAnotherThing(const MyOtherObject& obj);
void doYetAnotherThing(int value1, unsigned long value2, const std::string& value3, MyEnums::Seasons value4, const std::vector<MySecondObject>& value5);
std::string m_controllerType;
std::string m_networkType;
//...
};
//...
Config::Config(const std::string& cType, const std::string& nType) :
m_controllerType(cType),
m_networkType(nType)
{
}
My motivations and general understand of the subject:
use const references in constructors and methods to avoid double-copying when passing objects.
simple types - pass by value; classes and structs - pass by const reference (or simple reference when I need to modify them)
force compiler to create default move constructor and move assignment so that It would be able to do it's fancy magic and simultaneously it allows to avoid writing boring ctor() : m_v1(std::move(v1)), m_v2(std::move(v2)), m_v3(std::move(v3)) {}.
if it performs badly, use libc and raw pointers, then wrap it at class and write a comment.
I have a strong feeling that by rules of thumb are flawed and simply incorrect.
After reading cppreference, Scott Mayers, C++ standard, Stroustrup and so on, I feel like: "Yea, I understand every word here, but it still doesn't make any sense'. The only thing I king of understood is that move semantics makes sense when my class contains non-copiable types, like std::mutex and std::unique_ptr.
I've seen a lot of code where people pass complex object by value, like large strings, vectors and custom classes - I believe this is where move semantics happen, but, again, how can you pass an object to a function by move? If I am correct, it would leave an object in a "kind-of-null-state", making it unusable.
So, the questionы are:
- How do I correctly decide between pass-by-value and pass-by-reference?
- Do I need to provide both copy and move constructors?
- Do I need to explicitly write move and copy constructors? May I use = default? My classes are mostly POD object so there is no complex login involved.
- When debugging, I can always write std::cout << "move\n"; or std::cout << "copy\n"; in constructors of my own classes, but how do I know what happens with classes from stdlib?
P.S. It may look like it is a cry out of desperation (it is), not a valid SO question. I simply don't know to formulate my problems better than this.
If it is a primitive type, pass by value. Locality of reference wins.
If you aren't going to store a copy of it, pass by value or const&.
If you want to store a copy of it, and it is very cheap to move and modestly expensive to copy, pass by value.
If something has a modest cost to move, and is a sink parameter, consider pass by rvalue reference. Users will be forced to std::move.
Consider providing a way for callers to emplace construct into the field in highly generic code, or where you need every ounce of performance
The Rule of 0/3/5 describes how you should handle copy assign/construct/destroy. Ideally you follow the rule of 0; copy/move/destruct is all =default in anything except resource management types. If you want to implement any of copy/move/destruct, you need to implement, =default or =delete every other one of the 5.
If you are only taking 1 argument to a setter, consider writing both the && and const& versions of the setter. Or just exposing the underlying object. Move-assignment sometimes reuses storage and that is efficient.
Emplacing looks like this:
struct emplace_tag {};
struct wrap_foo {
template<class...Ts>
wrap_foo(emplace_tag, Ts&&...ts):
foo( std::forward<Ts>(ts)... )
{}
template<class T0, class...Ts>
wrap_foo(emplace_tag, std::initializer_list<T0> il, Ts&&...ts):
foo( il, std::forward<Ts>(ts)... )
{}
private:
Foo foo;
};
there are a myriad of other ways you can permit "emplace" construction. See emplace_back or emplace in standard containers as well (where they use placement ::new to construct objects, forwarding objects passed in).
Emplace construct even permits direct construction without even a move using objects with an operator T() setup properly. But that is something that is beyond the scope of this question.
I have a function that takes an std::optional
void foo(const std::optional<T>& opt);
But copying T is expencive.
Does this create a copy of T?
If so, how can I not create a copy?
As clarified by your follow up comment, you are calling the function foo like this:
T t;
...
foo({t});
In this case, the answer is yes. A copy of t will be created. To avoid that, you can use std::reference_wrapper to avoid copying:
void foo(const std::optional<std::reference_wrapper<const T>> &optT) {
...
}
With these changes to the function, calling the it in the same manner will not result in creation of a copy.
Yes, std::optional stores a copy of whatever you pass to it. The C++17 standard explicitly prohibits storing references in std::optional. See: https://en.cppreference.com/w/cpp/utility/optional#:~:text=There%20are%20no%20optional%20references
As suggested by others, passing std::optional<std::reference_wrapper<const T>> is one way to avoid making copies.
void foo(const std::optional<std::reference_wrapper<const T>>& opt) {
if (opt) {
// Do something.
}
}
T t;
foo(t);
But consider that the C++ committee had good reasons for disallowing references in std::optional. For instance, an "optional reference" essentially describes what a plain pointer does, and plain pointers don't suffer from the long type name.
void foo(const T* opt) {
if (opt) {
// Do something.
}
}
T t;
foo(&t);
I think you will find useful this documentation page. See the section about "Optional function parameters".
When you call the function and pass an instance of T, an optional will be constructed which will own its own copy of T, therefore it will call T's copy constructor.
int main()
{
T t;
optional<T> ot;
foo(t); // will create a copy
foo(ot); // won't create a copy
}
If you're using boost::optional, not std::optional, you can declare foo as receiving an optional reference, i.e.
void foo(boost::optional<const T&> t)
About std::move, here is what I can interpret, according to http://en.cppreference.com/w/cpp/utility/move :-
If I want to transfer ownership, I have to call std::move (or in rare case, std::forward).
Responsibility of std::move is calling operator=(A&& other).
The most essential step of the move operation is supposed to be implemented in operator=(A&&).
It is tricky to ensure that operator=(A&&) would be called. It need a special converter.
There are only two converters in the C++ world that can convert variables into xvalue (the &&) : std::move and std::forward.
Question
After adding many of std::move(std::unique_ptr) in my code, I start to worry that for such basic feature like transfer ownership, I have to heavily rely on the standard library (std::).
Do I really have to use std::move to transfer ownership?
Is spamming and hard-code calling std::move in many places of code-base a correct way to go for a high-standard program?
Should std::move be encapsulated?
They are actually a single question, but ask in different perspectives.
Edit
As request, here is my trial & error. It compiled ok.
I have no problem about the code, but I worry about its approach / pattern.
https://ideone.com/y8Pcgf
class T{
public: int value;
public: T(int a=1234){
value = a;
}
};
int main() {
std::unique_ptr<T> t1 = std::unique_ptr<T>(new T(1));
void* databaseNew=operator new [](sizeof(std::unique_ptr<T>));
std::unique_ptr<T>* t1ptr=static_cast<std::unique_ptr<T>*>(databaseNew);
new (t1ptr) std::unique_ptr<T>(std::move(t1));
return 0;
}
Rule of thumb:
If you're in a deduced x-value context, use std::forward:
template<class T>
void foo(T&& t) // T is deduced x-value, so we forward it
{
bar(std::forward<T>(t));
}
Otherwise use std::move
template<class T>
void foo1(std::vector<T> v) // although vector<T> is deduced, it's not an x-value
{
bar(std::move(v)); // so move it
}
template<class T>
void foo2(std::vector<T>&& v) // although vector<T> is deduced, it's not an x-value.
// In this case an r-value reference
{
bar(std::move(v)); // so move it
}
template<class T>
void foo3(std::vector<T>& v) // although vector<T> is deduced, it's not an x-value.
// In this case an l-value reference
{
bar(std::move(v)); // so move it
}
void foo4(std::vector<int> v) // complete type
{
bar(std::move(v)); // so move it
}
void foo5(std::vector<int> const & v) // const reference
{
bar(v); // not much point in moving it. std::move would cast it
// to std::vector<int> const&&, which although is detectable
// decays to std::vector<int> const&
}
which although is detectable... what?
It is permissible, if not necessarily advisable to write code like this:
#include <iostream>
struct X
{
void foo() const &
{
// do one thing...
std::cout << "one thing\n";
}
void foo() const &&
{
// do something else...
std::cout << "or another\n";
}
};
int main()
{
const X x;
x.foo();
std::move(x).foo();
}
const r-value references do exist, it's just that no-one uses them because there is no reasonable use-case.
The need to explicitly move, of which you complain, was actually done on purpose. Before unique_ptr, STL had a horrid construct called auto_ptr. It would move ownership impllicitly, and was borderline unusable unless you really really really knew what you were doing.
To make things more usable, in most cases C++ now requires you to explicitly state that you intend on moving ownership over a container, by using std::move.
In fact, std::move is little more than a cast to an rvalue reference.
There are cases where such an explicit specification is not necessary. For example, if the container from which you take ownership is already an rvalue (e.g. - a temporary object), then no case using std::move is necessary. For example, the following doesn't compile:
std::unique_ptr<int> a;
a = new int;
But the following does, without needing a move:
std::unique_ptr<int> a;
a = std::unique_ptr<int>(new int);
The reason this does not need a call to std::move, despite invoking the move operator, is that the object we move the ownership away from is already a temporary object (i.e. - an rvalue), so no cast is necessary.
Another example is if you call a function that returns a unique_ptr. You might have to call std::move inside the function to get it into the return value, but you do not need to call std::move on the function's return value to get it into the outside unique_ptr. It is already an rvalue, and therefor no cast is necessary.
Sometimes we like to take a large parameter by reference, and also to make the reference const if possible to advertize that it is an input parameter. But by making the reference const, the compiler then allows itself to convert data if it's of the wrong type. This means it's not as efficient, but more worrying is the fact that I think I am referring to the original data; perhaps I will take it's address, not realizing that I am, in effect, taking the address of a temporary.
The call to bar in this code fails. This is desirable, because the reference is not of the correct type. The call to bar_const is also of the wrong type, but it silently compiles. This is undesirable for me.
#include<vector>
using namespace std;
int vi;
void foo(int &) { }
void bar(long &) { }
void bar_const(const long &) { }
int main() {
foo(vi);
// bar(vi); // compiler error, as expected/desired
bar_const(vi);
}
What's the safest way to pass a lightweight, read-only reference? I'm tempted to create a new reference-like template.
(Obviously, int and long are very small types. But I have been caught out with larger structures which can be converted to each other. I don't want this to silently happen when I'm taking a const reference. Sometimes, marking the constructors as explicit helps, but that is not ideal)
Update: I imagine a system like the following: Imagine having two functions X byVal(); and X& byRef(); and the following block of code:
X x;
const_lvalue_ref<X> a = x; // I want this to compile
const_lvalue_ref<X> b = byVal(); // I want this to fail at compile time
const_lvalue_ref<X> c = byRef(); // I want this to compile
That example is based on local variables, but I want it to also work with parameters. I want to get some sort of error message if I'm accidentally passing a ref-to-temporary or a ref-to-a-copy when I think I'll passing something lightweight such as a ref-to-lvalue. This is just a 'coding standard' thing - if I actually want to allow passing a ref to a temporary, then I'll use a straightforward const X&. (I'm finding this piece on Boost's FOREACH to be quite useful.)
Well, if your "large parameter" is a class, the first thing to do is ensure that you mark any single parameter constructors explicit (apart from the copy constructor):
class BigType
{
public:
explicit BigType(int);
};
This applies to constructors which have default parameters which could potentially be called with a single argument, also.
Then it won't be automatically converted to since there are no implicit constructors for the compiler to use to do the conversion. You probably don't have any global conversion operators which make that type, but if you do, then
If that doesn't work for you, you could use some template magic, like:
template <typename T>
void func(const T &); // causes an undefined reference at link time.
template <>
void func(const BigType &v)
{
// use v.
}
If you can use C++11 (or parts thereof), this is easy:
void f(BigObject const& bo){
// ...
}
void f(BigObject&&) = delete; // or just undefined
Live example on Ideone.
This will work, because binding to an rvalue ref is preferred over binding to a reference-to-const for a temporary object.
You can also exploit the fact that only a single user-defined conversion is allowed in an implicit conversion sequence:
struct BigObjWrapper{
BigObjWrapper(BigObject const& o)
: object(o) {}
BigObject const& object;
};
void f(BigObjWrapper wrap){
BigObject const& bo = wrap.object;
// ...
}
Live example on Ideone.
This is pretty simple to solve: stop taking values by reference. If you want to ensure that a parameter is addressable, then make it an address:
void bar_const(const long *) { }
That way, the user must pass a pointer. And you can't get a pointer to a temporary (unless the user is being terribly malicious).
That being said, I think your thinking on this matter is... wrongheaded. It comes down to this point.
perhaps I will take it's address, not realizing that I am, in effect, taking the address of a temporary.
Taking the address of a const& that happens to be a temporary is actually fine. The problem is that you cannot store it long-term. Nor can you transfer ownership of it. After all, you got a const reference.
And that's part of the problem. If you take a const&, your interface is saying, "I'm allowed to use this object, but I do not own it, nor can I give ownership to someone else." Since you do not own the object, you cannot store it long-term. This is what const& means.
Taking a const* instead can be problematic. Why? Because you don't know where that pointer came from. Who owns this pointer? const& has a number of syntactic safeguards to prevent you from doing bad things (so long as you don't take its address). const* has nothing; you can copy that pointer to your heart's content. Your interface says nothing about whether you are allowed to own the object or transfer ownership to others.
This ambiguity is why C++11 has smart pointers like unique_ptr and shared_ptr. These pointers can describe real memory ownership relations.
If your function takes a unique_ptr by value, then you now own that object. If it takes a shared_ptr, then you now share ownership of that object. There are syntactic guarantees in place that ensure ownership (again, unless you take unpleasant steps).
In the event of your not using C++11, you should use Boost smart pointers to achieve similar effects.
You can't, and even if you could, it probably wouldn't help much.
Consider:
void another(long const& l)
{
bar_const(l);
}
Even if you could somehow prevent the binding to a temporary as input to
bar_const, functions like another could be called with the reference
bound to a temporary, and you'd end up in the same situation.
If you can't accept a temporary, you'll need to use a reference to a
non-const, or a pointer:
void bar_const(long const* l);
requires an lvalue to initialize it. Of course, a function like
void another(long const& l)
{
bar_const(&l);
}
will still cause problems. But if you globally adopt the convention to
use a pointer if object lifetime must extend beyond the end of the call,
then hopefully the author of another will think about why he's taking
the address, and avoid it.
I think your example with int and long is a bit of a red herring as in canonical C++ you will never pass builtin types by const reference anyway: You pass them by value or by non-const reference.
So let's assume instead that you have a large user defined class. In this case, if it's creating temporaries for you then that means you created implicit conversions for that class. All you have to do is mark all converting constructors (those that can be called with a single parameter) as explicit and the compiler will prevent those temporaries from being created automatically. For example:
class Foo
{
explicit Foo(int bar) { }
};
(Answering my own question thanks to this great answer on another question I asked. Thanks #hvd.)
In short, marking a function parameter as volatile means that it cannot be bound to an rvalue. (Can anybody nail down a standard quote for that? Temporaries can be bound to const&, but not to const volatile & apparently. This is what I get on g++-4.6.1. (Extra: see this extended comment stream for some gory details that are way over my head :-) ))
void foo( const volatile Input & input, Output & output) {
}
foo(input, output); // compiles. good
foo(get_input_as_value(), output); // compile failure, as desired.
But, you don't actually want the parameters to be volatile. So I've written a small wrapper to const_cast the volatile away. So the signature of foo becomes this instead:
void foo( const_lvalue<Input> input, Output & output) {
}
where the wrapper is:
template<typename T>
struct const_lvalue {
const T * t;
const_lvalue(const volatile T & t_) : t(const_cast<const T*>(&t_)) {}
const T* operator-> () const { return t; }
};
This can be created from an lvalue only
Any downsides? It might mean that I accidentally misuse an object that is truly volatile, but then again I've never used volatile before in my life. So this is the right solution for me, I think.
I hope to get in the habit of doing this with all suitable parameters by default.
Demo on ideone