I'm trying to find a good way to write a setter function for a template class. For non-template classes if somehow trivial because the function signature/implementation depends on parameter type. For example if the parameter type is int the next function should be optimal:
void MyClass::Set(int value)
{
myValue = value;
}
If the parameter type is std::vector next implementation should be close to optimal:
void MyClass::Set(std::vector<SomeType> value)
{
std::swap(myValue, value);
}
as the right constructor (move or copy) will be used to construct the function parameter and no unnecessary copying occurs, assuming move constructor cost is negligible.
As you can see both implementations have drawbacks when the type is changed: If the type is changed to std::vector for the first version, at least one unnecessary copy is made increasing the actual cost by a factor of 2 or 3.
If the type changed to int in the second version 2 unnecessary copies are made, increasing the actual cost by a factor of 3.
Can you please give me a good generic implementation for a setter function (maybe with overloads)? It should be optimal/close to optimal for any type used as parameter.
PS: I would prefer to not use std::enable_if to make several type dependent setters as the class will increase dramatically.
You can use a forwarding reference to accept either an rvalue or an lvalue, then forward that to the appropriate move or copy assignment operator:
template <typename T>
void Set(T && value) {
myValue = std::forward<T>(value);
}
You should have 3 overloads.
void Set( T&& t ) {
v = std::move(t);
}
void Set( T const& t ) {
v = t;
}
template<class U>
void Set( U&& u ) {
v = std::forward<U>(u);
}
the first two allow implicit conversion to work well, if the caller uses {} initialization on the argument, or if the caller is the name of a function and it needs a context, or a few other cases: it will work. This fixes some of the bigger annoyances with "perfect forwarding".
The third gives you anything for which your operator= is overloaded to handle that isn't covered by the first two. It is an example of "perfect forwarding", which as the name implies is imperfect.
The most common problem if you only use the 3rd overload, .Set( {construction args} ) doesn't work.
Possibly the const& overload is redundant, but I am unsure.
live example
// and sometimes:
template<class...Us>
void Set( Us&&...us ) {
v = {std::forward<Us>(us)...};
}
The forth is an emplace-set. This is not actually useful in this case (where we are using operator=), but in some cases it can be. For an example of where such an emplace-accessor is useful, optional::value_or should have that overload. Basically, cases where you could directly construct the target value, or where other parameters might lead to you not constructing the value, this is useful: if you do a {std::forward<Us>(us)...} always you might as well use .Set( {args...} ) externally, which calls the first overload above.
Related
I have this class:
template<typename T, size_t N>
class Array {
private:
T array[N];
public:
template <typename... InitValues>
constexpr Array(InitValues... init_values)
: array{ init_values... } {}
[[nodiscard]]
consteval int len() const noexcept { return sizeof(array) / sizeof(T); }
}
I would like to know, for such a simple member function, when I should provide the necessary ref-qualified overloads.
With the actual code, I can compile and run the following code:
constexpr collections::Array a = collections::Array<long, 5>{1L, 2L, 3L};
SECTION("length of the array") {
REQUIRE( a.len() == 5 );
REQUIRE( collections::Array<int, 1>{1}.len() == 1 );
}
1- Why I can compile the second REQUIRE that contains the call with the rvalue?
Now I am gonna change the len() member function to this:
[[nodiscard]]
consteval int len() const& noexcept { return sizeof(array) / sizeof(T); }
2- Why I can compile both with the const&? I suppose that they are two are different ref-qualified usages. I assume that I can make the call with the first one, which is an lvalue, but can't understand why I can compile the second having defined the len() method as const&.
Last change:
[[nodiscard]]
consteval int len() const&& noexcept { return sizeof(array) / sizeof(T); }
And finally, I got a compiler error on a.get<I>().
'this' argument to member function 'len' is an lvalue, but function has rvalue ref-qualifier
REQUIRE( a.len() == 5 );
that works perfect if I comment that line of code and I just run:
REQUIRE( collections::Array<int, 1>{1}.len() == 1 );
and also I could use std::move(a) to perform the cast of a to an rvalue reference and make the code compile. But I don't want to do that.
What is the correct way of code those examples in terms of ref-qualified overloads?
Don't forget about the questions on the examples above
EDIT:
I will add another member function that could potentially do different things based on the ref-qualified implementation (or that what I am suppose that could happen):
template <size_t I>
requires concepts::AccessInBounds<I, N>
constexpr T get() const noexcept {
return array[I];
}
template <size_t I>
requires concepts::AccessInBounds<I, N>
constexpr T& get() const& noexcept {
return array[I];
}
To question 1: why not? The rule is the same as for lvalues: you can call const member functions regardless of the constness of the object.
To question 2: Because it is meant to be identical to having a const& function parameter: the function can be called with any lvalue or rvalue. It exists primarily to allow you to distinguish between lvalue and rvalue overloads:
class Array {
// These two declarations would be ambiguous for Array rvalues
// int len() const;
// int len() &&;
// These are not: your test expressions will use different overloads
int len() const&;
int len() &&;
};
The two functions in your edit are also ambiguous, for both lvalues and rvalues. A motivating example would be more along these lines: suppose my class provides functionality to some resource that could be expensive to copy, but is cheaper to move, say a std::vector.
template<class T>
class VectorView {
std::vector<T> vector;
public:
// ...
constexpr std::vector<T> const& base() const noexcept { return vector; }
};
Now there is no way for a user of this class to transfer ownership of the vector data back from a view object, even if that would be useful when calling the base() function on an rvalue. Because it is in the spirit of C++ to avoid paying for things you do not need, you could allow this by adding an rvalue-qualified overload that instead returns an rvalue reference using std::move.
So the answer to whether you need this kind of overload is it depends, which is unfortunately also in the spirit of C++. If you were implementing something like my example class for the standard library, then you certainly would, because it is based on std::ranges::owning_view. As you can see on that page, it covers all four possible base()s. If you were instead only using a reference to a source range, it would be unexpected and inappropriate to move from that object, so the related ref_view only has a const base() function like the one I wrote.
Edit As for move semantics, the difference between something like an array and a vector is that Array<T,N> is based on T[N], while std::vector<T> is based on T*. Moving the array requires N move operations (linear time complexity), and whether a move is an improvement over a copy depends on T. Also, it needs memory space for 2N elements. On the other hand, a vector only ever needs three pointers to do its job, so it can be moved in constant time, while copying still takes linear time.
This potential gain is the rationale for move semantics and rvalue references in a nutshell. The ability to also have &&-qualified member functions completes this language feature, but is not as significant as move constructors and assignment functions. I also found the answers to this question useful, as they give some more examples of ref-qualified overloads.
My question is what is the advantage of perfect forwarding a function, which is passed to our wrapper.
template<typename T, typename ...U>
auto time_function(T&& func, U&& ...args)
{
std::cout<<"timing void function"<<std::endl;
//std::forward<T>(func)(std::forward<U>(args)...); //this vs next one
func(std::forward<U>(args)...);
std::cout<<"timing over"<<std::endl;
}
In the case of argument forwarding, it is clear that forwarding preserves the lvalueness vs rvalueness of the argument. However, is there any use to forward func before calling?
Let's say I pass both temporary functors and normal functions to the time_function wrapper.
Suppose that I have a stateful assignor functor. The assignor will be reused multiple times, so it has to copy the value each time. However, if the assignor is an rvalue, it can just move the value:
struct Assignor {
std::string value;
void operator()(std::string& dest) const &
{
dest = value;
}
void operator()(std::string& dest) &&
{
dest = std::move(value);
}
};
Now, perfect forwarding makes a difference on rvalues:
Assignor f{std::string(10000, 'X')};
std::string a, b, c;
time_function(f, a); // copies the string
time_function(std::move(f), b); // should move the string
// but copies if you don't std::forward
time_function(Assignor{std::string(10000, 'Y')}, c); // same
(This is just an example of how a functor can be optimized w.r.t. value category. I know it looks a bit artificial, but people always come up with creative ideas.)
By the way, you should be using std::invoke instead of directly calling ():
std::invoke(std::forward<T>(func), std::forward<U>(args)...);
In addition to L.F.'s answer I want to make a small note about a potential problem, that is obvious but can still be overlooked: sometimes it is dangerous to pass functional objects by a universal reference and invoke them as rvalues. Suppose that instead of
void time_function(T&& func, ...)
that invokes std::forward<T>(func)(...) once, you have
void for_each(T&& func, ...)
that potentially invokes std::forward<T>(func)(...) multiple times. Then after the first call, all further calls are not safe. In the L.F.'s example, this corresponds to multiple assignments from a "moved-from" state: after the first assignment, std::string value member will be left in a "valid but unspecified state", and later assignments won't do what they are expected to do (though they won't be UB). If the Assignor doesn't have the && overload of operator(), the problem won't show up.
I'm used to pass lambda functions (and other callables) to template functions -- and use them -- as follows
template <typename F>
auto foo (F && f)
{
// ...
auto x = std::forward<F>(f)(/* some arguments */);
// ...
}
I mean: I'm used to pass them through a forwarding reference and call them passing through std::forward.
Another Stack Overflow user argue (see comments to this answer) that this, calling the functional two or more time, it's dangerous because it's semantically invalid and potentially dangerous (and maybe also Undefined Behavior) when the function is called with a r-value reference.
I've partially misunderstand what he means (my fault) but the remaining doubt is if the following bar() function (with an indubitable multiple std::forward over the same object) it's correct code or it's (maybe only potentially) dangerous.
template <typename F>
auto bar (F && f)
{
using A = typename decltype(std::function{std::forward<F>(f)})::result_type;
std::vector<A> vect;
for ( auto i { 0u }; i < 10u ; ++i )
vect.push_back(std::forward<F>(f)());
return vect;
}
Forward is just a conditional move.
Therefore, to forward the same thing multiple times is, generally speaking, as dangerous as moving from it multiple times.
Unevaluated forwards don't move anything, so those don't count.
Routing through std::function adds a wrinkle: that deduction only works on function pointers and on function objects with a single function call operator that is not && qualified. For these, rvalue and lvalue invocation are always equivalent if both compiles.
I'd say the general rule applies in this case. You're not supposed to do anything with a variable after it was moved/forwarded from, except maybe assigning to it.
Thus...
How do correctly use a callable passed through forwarding reference?
Only forward if you're sure it won't be called again (i.e. on last call, if at all).
If it's never called more than once, there is no reason to not forward.
As for why your snippet could be dangerous, consider following functor:
template <typename T>
struct foo
{
T value;
const T &operator()() const & {return value;}
T &&operator()() && {return std::move(value);}
};
As an optimization, operator() when called on an rvalue allows caller to move from value.
Now, your template wouldn't compile if given this functor (because, as T.C. said, std::function wouldn't be able to determine return type in this case).
But if we changed it a bit:
template <typename A, typename F>
auto bar (F && f)
{
std::vector<A> vect;
for ( auto i { 0u }; i < 10u ; ++i )
vect.push_back(std::forward<F>(f)());
return vect;
}
then it would break spectacularly when given this functor.
If you're either going to just forward the callable to another place or simply call the callable exactly once, I would argue that using std::forward is the correct thing to do in general. As explained here, this will sort of preserve the value category of the callable and allow the "correct" version of a potentially overloaded function call operator to be called.
The problem in the original thread was that the callable was being called in a loop, thus potentially invoked more than once. The concrete example from the other thread was
template <typename F>
auto map(F&& f) const
{
using output_element_type = decltype(f(std::declval<T>()));
auto sequence = std::make_unique<Sequence<output_element_type>>();
for (const T& element : *this)
sequence->push(f(element));
return sequence;
}
Here, I believe that calling std::forward<F>(f)(element) instead of f(element), i.e.,
template <typename F>
auto map(F&& f) const
{
using output_element_type = decltype(std::forward<F>(f)(std::declval<T>()));
auto sequence = std::make_unique<Sequence<output_element_type>>();
for (const T& element : *this)
sequence->push(std::forward<F>(f)(element));
return sequence;
}
would be potentially problematic. As far as my understanding goes, the defining characteristic of an rvalue is that it cannot explicitly be referred to. In particular, there is naturally no way for the same prvalue to be used in an expression more than once (at least I can't think of one). Furthermore, as far as my understanding goes, if you're using std::move or std::forward or whatever other way to obtain an xvalue, even on the same original object, the result will be a new xvalue every time. Thus, there also cannot possibly be a way to refer to the same xvalue more than once. Since the same rvalue cannot be used more than once, I would argue (see also comments underneath this answer) that it would generally be a valid thing for an overloaded function call operator to do something that can only be done once in case the call happens on an rvalue, for example:
class MyFancyCallable
{
public:
void operator () & { /* do some stuff */ }
void operator () && { /* do some stuff in a special way that can only be done once */ }
};
The implementation of MyFancyCallable may assume that a call that would pick the &&-qualified version cannot possibly happen more than once (on the given object). Thus, I would consider forwarding the same callable into more than one call to be semantically broken.
Of course, technically, there is no universal definition of what it actually means to forward or move an object. In the end, it's really up to the implementation of the particular types involved to assign meaning there. Thus, you may simply specify as part of your interface that potential callables passed to your algorithm must be able to deal with being called multiple times on an rvalue that refers to the same object. However, doing so pretty much goes against all the conventions for how the rvalue reference mechanism is generally used in C++, and I don't really see what there possibly would be to be gained from doing this…
I've recently been looking into forwarding references in C++ and below is a quick summary of my current understanding of the concept.
Let's say I have a template function footaking a forwarding reference to a single argument of type T.
template<typename T>
void foo(T&& arg);
If I call this function with an lvalue then T will be deduced as T& making the arg parameter be of type T& due to the reference collapsing rules T& && -> T&.
If this function gets called with an unnamed temporary, such as the result of a function call, then Twill be deduced as T making the arg parameter be of type T&&.
Inside foo however, arg is a named parameter so I will need to use std::forward if I want to pass the parameter along to other functions and still maintain its value category.
template<typename T>
void foo(T&& arg)
{
bar(std::forward<T>(arg));
}
As far as I understand the cv-qualifiers are unaffected by this forwarding. This means that if I call foo with a named const variable then T will be deduced as const T& and hence the type of arg will also be const T& due to the reference collapsing rules. For const rvalues T will be deduced as const T and hence arg will be of type const T&&.
This also means that if I modify the value of arg inside foo I will get a compile time error if I did infact pass a const variable to it.
Now onto my question.
Assume I am writing a container class and want to provide a method for inserting objects into my container.
template<typename T>
class Container
{
public:
void insert(T&& obj) { storage[size++] = std::forward<T>(obj); }
private:
T *storage;
std::size_t size;
/* ... */
};
By making the insert member function take a forwarding reference to obj I can use std::forward to take advantage of the move assignment operator of the stored type T if insert was infact passed a temporary object.
Previously, when I didn't know anything about forwarding references I would have written this member function taking a const lvalue reference:
void insert(const T& obj).
The downside of this is that this code does not take advantage of the (presumably more efficient) move assignment operator if insert was passed a temporary object.
Assuming I haven't missed anything.
Is there any reason to provide two overloads for the insert function? One taking a const lvalue reference and one taking a forwarding reference.
void insert(const T& obj);
void insert(T&& obj);
The reason I'm asking is that the reference documentation for std::vectorstates that the push_back method comes in two overloads.
void push_back (const value_type& val);
void push_back (value_type&& val);
Why is the first version (taking a const value_type&) needed?
You have to be careful about function templates, versus non-template methods of class templates. Your member insert is not itself a template. It's a method of a template class.
Container<int> c;
c.insert(...);
We can pretty easily see that T is not deduced on the second line, because it's already fixed to int on the first line, because T is a template parameter of the class, not the method.
Non-template methods of class templates, only differ from regular methods in one way, once the class has been instantiated: they aren't instantiated unless they are actually called. This is useful because it allows a template class to work with types, for which only some of the methods make sense (STL containers are full of examples like this).
The bottom line is that in my example above, since T is fixed to int, your method becomes:
void insert(int&& obj) { storage[size++] = std::forward<int>(obj); }
This is not a forwaring reference at all, but simply takes by rvalue reference, i.e. it only binds to rvalues. That is why you typically see two overloads for things like push_back, one for lvalues and one for rvalues.
#Nir Friedman already answered the question, so I'm going to offer some additional advice.
If your Container class is not meant to store polymorphic types (which is common of containers, including std::vector and other similar STL containers), you can get away with simplifying your code, in the way you're trying to do in your original example.
Instead of:
void insert(T const& t) {
storage[size++] = t;
}
void insert(T && t) {
storage[size++] = std::move(t);
}
You could get perfectly correct code by writing the following instead:
void insert(T t) {
storage[size++] = std::move(t);
}
The reason for this is that if the object is being copied in, t will be copy-constructed with the object provided, and then move-assigned into storage[size++], whereas if the object is being moved in, t will be move-constructed with the object provided, and then move-assigned into storage[size++]. So you've simplified your code at the cost of a single extra move-assignment, which many compilers will happily optimize out.
There is a major downside to this approach, though: If the object defines a copy-constructor and doesn't define a move-constructor (common for older types in legacy code), this results in double-copies in all cases. Your compiler might be able to optimize it away (because compilers can optimize to completely different code so long as the user-visible effects are unchanged), but maybe not. That could be a significant performance hit if you have to work with heavy objects that don't implement move-semantics. This is probably the reason STL containers don't use this technique (they value performance over brevity). But if you're looking for a way to reduce the amount of boilerplate code you write, and aren't worried about having to use "copy-only" objects, then this will probably work fine for you.
I have a library function (not under my control here) which take an r value reference to the movable and copyable type Bar:
void foo(Bar&& b);
In my own code, I sometimes need to give it a copy of an existing value, such as
const auto b = getBar();
foo(mk_copy(b));
b.baz();
This is what comes to mind,
template<typename T> auto mk_copy(T val){return val;}
Is there a standard way to do this, or a more common pattern? Perhaps even
template<typename T> auto mk_copy(T&& val){return std::forward<T>(val);}
Or, as pscill points out just writing the name of the class again,
foo(Bar(b));
but I prefer not to repeat type names.
For the built-in types the prefix + plays the role of “make a copy”:
void foo( char&& ) {}
void bar( double*&& ) {}
auto main() -> int
{
char x{};
foo( +x );
double* p{};
bar( +p );
}
Still, it would probably be confusing to readers of the code if you applied prefix + to some other type, and a general prefix operator+ template might end up in conflict with some class' own prefix operator+.
So, I suggest using the now apparent naming convention for makers, namely a make prefix.
Then it can look like this:
template< class Type >
auto make_copy( Type const& o ) -> Type { return o; }
Your first proposed solution,
template<typename T> auto mk_copy(T val){return val;}
suffers from potentially copying the value twice: first copying to the by-value argument, and then copying to the function result.
This is not a problem for a movable type such as a standard library container, because the return type copying will reduce to a move. But it can be a problem for largish non-movable type, as can occur in legacy code.
The second proposed solution,
template<typename T> auto mk_copy(T&& val){return std::forward<T>(val);}
takes the argument by forwarding reference (a.k.a. universal reference), and then deduces the return type from a re-creation of the argument type. The return type will always be a non-reference, since that's what plain auto deduces, so this is technically correct. But it's needlessly complicated.
There is no standard mechanism, but if you want one, your examples aren't very good. The first mk_copy copies T twice (or copies and moves). The second one seems very confusing as to what it's trying to do.
The obvious way is to simply take a const T&, like you normally would:
template<typename T> T mk_copy(const T &val){return val;}
I think you'd better just define that mk_copy method, or add one more declaration:
auto b_copy = b;
foo(std::move(b_copy));
These are clearer to the readers.
Just to demonstrate the possibilities, you could use decltype to get the type:
foo(std::decay_t<decltype(b)>{b});
or you could get the same effect with lambdas:
foo([&]{return b;}());
or tuples:
foo(std::get<0>(std::make_tuple(b)));
or pairs:
foo(std::make_pair(b, 0).first);