When I generate a new thread (std::thread) with a function the arguments of that function
are by value - not by reference.
So if I define that function with a reference argument (int& nArg)
my compiler (mingw 4.9.2) outputs an error (in compilian-suaeli something like
"missing copy constructor" I guess ;-)
But if I make that reference argument const (const int& nArg) it does not complain.
Can somebody explain please?
If you want to pass reference, you have to wrap it into std::reference_wrapper thanks to std::ref. Like:
#include <functional>
#include <thread>
void my_function(int&);
int main()
{
int my_var = 0;
std::thread t(&my_function, std::ref(my_var));
// ...
t.join();
}
std::thread's arguments are used once.
In effect, it stores them in a std::tuple<Ts...> tup. Then it does a f( std::get<Is>(std::move(tup))...).
Passing std::get an rvalue tuple means that it is free to take the state from a value or rvalue reference field in the tuple. Without the tuple being an rvalue, it instead gives a reference to it.
Unless you use reference_wrapper (ie, std::ref/std::cref), the values you pass to std::thread are stored as values in the std::tuple. Which means the function you call is passed an rvalue to the value in the std::tuple.
rvalues can bind to const& but not to &.
Now, the std::tuple above is an implementation detail, an imagined implementation of std::thread. The wording in the standard is more obtuse.
Why does the standard say this happens? In general, you should not bind a & parameter to a value which will be immediately discarded. The function thinks that it is modifying something that the caller can see; if the value will be immediately discarded, this is usually an error on the part of the caller.
const& parameters, on the other hand, do bind to values that will be immediately discarded, because we use them for efficiency purposes not just for reference purposes.
Or, roughly, because
const int& x = 7;
is legal
int& x = 7;
is not. The first is a const& to a logically discarded object (it isn't due to reference lifetime extension, but it is logically a temporary).
Related
I've noticed that it's impossible to pass a non-const reference as an argument to std::async.
#include <functional>
#include <future>
void foo(int& value) {}
int main() {
int value = 23;
std::async(foo, value);
}
My compiler (GCC 4.8.1) gives the following error for this example:
error: no type named ‘type’ in ‘class std::result_of<void (*(int))(int&)>’
But if I wrap the value passed to std::async in std::reference_wrapper, everything is OK. I assume this is because std::async takes it's arguments by value, but I still don't understand the reason for the error.
It's a deliberate design choice/trade-off.
First, it's not necessarily possible to find out whether the functionoid passed to async takes its arguments by reference or not. (If it's not a simple function but a function object, it could have an overloaded function call operator, for example.) So async cannot say, "Hey, let me just check what the target function wants, and I'll do the right thing."
So the design question is, does it take all arguments by reference if possible (i.e. if they're lvalues), or does it always make copies? Making copies is the safe choice here: a copy cannot become dangling, and a copy cannot exhibit race conditions (unless it's really weird). So that's the choice that was made: all arguments are copied by default.
But then, the mechanism is written so that it actually fails to then pass the arguments to a non-const lvalue reference parameter. That's another choice for safety: otherwise, the function that you would expect to modify your original lvalue instead modifies the copy, leading to bugs that are very hard to track down.
But what if you really, really want the non-const lvalue reference parameter? What if you promise to watch out for dangling references and race conditions? That's what std::ref is for. It's an explicit opt-in to the dangerous reference semantics. It's your way of saying, "I know what I'm doing here."
std::async (and other functions that do perfect forwarding) look at the type of the argument that you pass to figure out what to do. They do not look at how that argument will eventually be used. So, to pass an object by reference you need to tell std::async that you're using a reference. However, simply passing a reference won't do that. You have to use std::ref(value) to pass value by reference.
The issue itself is only marginally related to std::async(): When defining the result of the operation, std::async() uses std::result_of<...>::type with all its arguments being std::decay<...>::type'ed. This is reasonable because std::async() takes arbitrary types and forwards them to store them in some location. To store them, values are needed for the function object as well as for the arguments. Thus, std::result_of<...> is used similar to this:
typedef std::result_of<void (*(int))(int&)>::type result_type;
... and since int can't be bound to an int& (int isn't an lvalue type was is needed to be bound to int&), this fails. Failure in this case means that std::result_of<...> doesn't define a nested type.
A follow-up question could be: What is this type used to instantiate std::result_of<...>? The idea is that the function call syntax consisting of ResultType(ArgumentTypes...) is abused: instead of a result type, a function type is passed and std::result_of<...> determines the type of the function called when that function type is called with the given list of arguments is called. For function pointer types it isn't really that interesting but the function type can also be a function object where overloading needs to be taken into account. So basically, std::result_of<...> is used like this:
typedef void (*function_type)(int&);
typedef std::result_of<function_type(int)>::type result_type; // fails
typedef std::result_of<function_type(std::reference_wrapper<int>)>::type result_type; //OK
if I have a function:
Foo& Bar()
{
return /// do something to create a non-temp Foo here and return a reference to it
}
why is this:
auto x = Bar(); /// probably calls copy ctor - haven't checked
not the same as this?
auto &x = Bar(); /// actually get a reference here
(Actually, I'd expect the second version to get a reference to a reference, which makes little sense.)
If I explicitly specified the type of x as a value or a reference, I'll get what I expect (of course). I would expect, though, that auto would compile to the return type of Bar(), which, in this case, is a reference.
Is there an implicit cast between Foo and Foo& that comes into play here?
(Spec references accepted, though I'm getting tired of reading committee-speak.)
(Second use of time machine will be making C++ pass by reference by default. With a #pragma compatibility trigger for compiling C code. ARGH.)
The type deduction for auto works exactly the same as for templates:
when you deduce auto you will get a value type.
when you deduce auto& you wil get a non-const reference type
when you deduce const auto& you will get a const reference
when you deduce auto&& you will get
a non-const reference if you assign a non-const reference
a const reference if you assign a const reference
a value when you assign a temporary
Taken directly from Herb Sutter's blog post:
auto means “take exactly the type on the right-hand side, but strip off top-level const/volatile and &/&&.”
if I have a function:
Foo& Bar()
{
return /// do something to create a non-temp Foo here and return a reference to it
}
why is this:
auto x = Bar(); /// probably calls copy ctor - haven't checked
not the same as this?
auto &x = Bar(); /// actually get a reference here
(Actually, I'd expect the second version to get a reference to a reference, which makes little sense.)
If I explicitly specified the type of x as a value or a reference, I'll get what I expect (of course). I would expect, though, that auto would compile to the return type of Bar(), which, in this case, is a reference.
Is there an implicit cast between Foo and Foo& that comes into play here?
(Spec references accepted, though I'm getting tired of reading committee-speak.)
(Second use of time machine will be making C++ pass by reference by default. With a #pragma compatibility trigger for compiling C code. ARGH.)
The type deduction for auto works exactly the same as for templates:
when you deduce auto you will get a value type.
when you deduce auto& you wil get a non-const reference type
when you deduce const auto& you will get a const reference
when you deduce auto&& you will get
a non-const reference if you assign a non-const reference
a const reference if you assign a const reference
a value when you assign a temporary
Taken directly from Herb Sutter's blog post:
auto means “take exactly the type on the right-hand side, but strip off top-level const/volatile and &/&&.”
For move enabled classes is there a difference between this two?
struct Foo {
typedef std::vector<std::string> Vectype;
Vectype m_vec;
//this or
void bar(Vectype&& vec)
{
m_vec = std::move(vec);
}
//that
void bar(Vectype vec)
{
m_vec = std::move(vec);
}
};
int main()
{
Vectype myvec{"alpha","beta","gamma"};
Foo fool;
fool.bar(std::move(myvec));
}
My understanding is that if you use a lvalue myvec you also required to introduce const
Vectype& version of Foo::bar() since Vectype&& won't bind. That's aside, in the rvalue case, Foo::bar(Vectype) will construct the vector using the move constructor or better yet elide the copy all together seeing vec is an rvalue (would it?). So is there a compelling reason to not to prefer by value declaration instead of lvalue and rvalue overloads?
(Consider I need to copy the vector to the member variable in any case.)
The pass-by-value version allows an lvalue argument and makes a copy of it. The rvalue-reference version can't be called with an lvalue argument.
Use const Type& when you don't need to change or copy the argument at all, use pass-by-value when you want a modifiable value but don't care how you get it, and use Type& and Type&& overloads when you want something slightly different to happen depending on the context.
The pass-by-value function is sufficient (and equivalent), as long as the argument type has an efficient move constructor, which is true in this case for std::vector.
Otherwise, using the pass-by-value function may introduce an extra copy-construction compared to using the pass-by-rvalue-ref function.
See the answer https://stackoverflow.com/a/7587151/1190077 to the related question Do I need to overload methods accepting const lvalue reference for rvalue references explicitly? .
Yes, the first one (Vectype&& vec) won't accept a const object or simply lvalue.
If you want to save the object inside like you do, it's best to copy(or move if you pass an rvalue) in the interface and then move, just like you did in your second example.
If you read code like
auto&& var = foo();
where foo is any function returning by value of type T. Then var is an lvalue of type rvalue reference to T. But what does this imply for var? Does it mean, we are allowed to steal the resources of var? Are there any reasonable situations when you should use auto&& to tell the reader of your code something like you do when you return a unique_ptr<> to tell that you have exclusive ownership? And what about for example T&& when T is of class type?
I just want to understand, if there are any other use cases of auto&& than those in template programming; like the ones discussed in the examples in this article Universal References by Scott Meyers.
By using auto&& var = <initializer> you are saying: I will accept any initializer regardless of whether it is an lvalue or rvalue expression and I will preserve its constness. This is typically used for forwarding (usually with T&&). The reason this works is because a "universal reference", auto&& or T&&, will bind to anything.
You might say, well why not just use a const auto& because that will also bind to anything? The problem with using a const reference is that it's const! You won't be able to later bind it to any non-const references or invoke any member functions that are not marked const.
As an example, imagine that you want to get a std::vector, take an iterator to its first element and modify the value pointed to by that iterator in some way:
auto&& vec = some_expression_that_may_be_rvalue_or_lvalue;
auto i = std::begin(vec);
(*i)++;
This code will compile just fine regardless of the initializer expression. The alternatives to auto&& fail in the following ways:
auto => will copy the vector, but we wanted a reference
auto& => will only bind to modifiable lvalues
const auto& => will bind to anything but make it const, giving us const_iterator
const auto&& => will bind only to rvalues
So for this, auto&& works perfectly! An example of using auto&& like this is in a range-based for loop. See my other question for more details.
If you then use std::forward on your auto&& reference to preserve the fact that it was originally either an lvalue or an rvalue, your code says: Now that I've got your object from either an lvalue or rvalue expression, I want to preserve whichever valueness it originally had so I can use it most efficiently - this might invalidate it. As in:
auto&& var = some_expression_that_may_be_rvalue_or_lvalue;
// var was initialized with either an lvalue or rvalue, but var itself
// is an lvalue because named rvalues are lvalues
use_it_elsewhere(std::forward<decltype(var)>(var));
This allows use_it_elsewhere to rip its guts out for the sake of performance (avoiding copies) when the original initializer was a modifiable rvalue.
What does this mean as to whether we can or when we can steal resources from var? Well since the auto&& will bind to anything, we cannot possibly try to rip out vars guts ourselves - it may very well be an lvalue or even const. We can however std::forward it to other functions that may totally ravage its insides. As soon as we do this, we should consider var to be in an invalid state.
Now let's apply this to the case of auto&& var = foo();, as given in your question, where foo returns a T by value. In this case we know for sure that the type of var will be deduced as T&&. Since we know for certain that it's an rvalue, we don't need std::forward's permission to steal its resources. In this specific case, knowing that foo returns by value, the reader should just read it as: I'm taking an rvalue reference to the temporary returned from foo, so I can happily move from it.
As an addendum, I think it's worth mentioning when an expression like some_expression_that_may_be_rvalue_or_lvalue might turn up, other than a "well your code might change" situation. So here's a contrived example:
std::vector<int> global_vec{1, 2, 3, 4};
template <typename T>
T get_vector()
{
return global_vec;
}
template <typename T>
void foo()
{
auto&& vec = get_vector<T>();
auto i = std::begin(vec);
(*i)++;
std::cout << vec[0] << std::endl;
}
Here, get_vector<T>() is that lovely expression that could be either an lvalue or rvalue depending on the generic type T. We essentially change the return type of get_vector through the template parameter of foo.
When we call foo<std::vector<int>>, get_vector will return global_vec by value, which gives an rvalue expression. Alternatively, when we call foo<std::vector<int>&>, get_vector will return global_vec by reference, resulting in an lvalue expression.
If we do:
foo<std::vector<int>>();
std::cout << global_vec[0] << std::endl;
foo<std::vector<int>&>();
std::cout << global_vec[0] << std::endl;
We get the following output, as expected:
2
1
2
2
If you were to change the auto&& in the code to any of auto, auto&, const auto&, or const auto&& then we won't get the result we want.
An alternative way to change program logic based on whether your auto&& reference is initialised with an lvalue or rvalue expression is to use type traits:
if (std::is_lvalue_reference<decltype(var)>::value) {
// var was initialised with an lvalue expression
} else if (std::is_rvalue_reference<decltype(var)>::value) {
// var was initialised with an rvalue expression
}
First, I recommend reading this answer of mine as a side-read for a step-by-step explanation on how template argument deduction for universal references works.
Does it mean, we are allowed to steal the resources of var?
Not necessarily. What if foo() all of a sudden returned a reference, or you changed the call but forgot to update the use of var? Or if you're in generic code and the return type of foo() might change depending on your parameters?
Think of auto&& to be exactly the same as the T&& in template<class T> void f(T&& v);, because it's (nearly†) exactly that. What do you do with universal references in functions, when you need to pass them along or use them in any way? You use std::forward<T>(v) to get the original value category back. If it was an lvalue before being passed to your function, it stays an lvalue after being passed through std::forward. If it was an rvalue, it will become an rvalue again (remember, a named rvalue reference is an lvalue).
So, how do you use var correctly in a generic fashion? Use std::forward<decltype(var)>(var). This will work exactly the same as the std::forward<T>(v) in the function template above. If var is a T&&, you'll get an rvalue back, and if it is T&, you'll get an lvalue back.
So, back on topic: What do auto&& v = f(); and std::forward<decltype(v)>(v) in a codebase tell us? They tell us that v will be acquired and passed on in the most efficient way. Remember, though, that after having forwarded such a variable, it's possible that it's moved-from, so it'd be incorrect use it further without resetting it.
Personally, I use auto&& in generic code when I need a modifyable variable. Perfect-forwarding an rvalue is modifying, since the move operation potentially steals its guts. If I just want to be lazy (i.e., not spell the type name even if I know it) and don't need to modify (e.g., when just printing elements of a range), I'll stick to auto const&.
† auto is in so far different that auto v = {1,2,3}; will make v an std::initializer_list, whilst f({1,2,3}) will be a deduction failure.
Consider some type T which has a move constructor, and assume
T t( foo() );
uses that move constructor.
Now, let's use an intermediate reference to capture the return from foo:
auto const &ref = foo();
this rules out use of the move constructor, so the return value will have to be copied instead of moved (even if we use std::move here, we can't actually move through a const ref)
T t(std::move(ref)); // invokes T::T(T const&)
However, if we use
auto &&rvref = foo();
// ...
T t(std::move(rvref)); // invokes T::T(T &&)
the move constructor is still available.
And to address your other questions:
... Are there any reasonable situations when you should use auto&& to tell the reader of your code something ...
The first thing, as Xeo says, is essentially I'm passing X as efficiently as possible, whatever type X is. So, seeing code which uses auto&& internally should communicate that it will use move semantics internally where appropriate.
... like you do when you return a unique_ptr<> to tell that you have exclusive ownership ...
When a function template takes an argument of type T&&, it's saying it may move the object you pass in. Returning unique_ptr explicitly gives ownership to the caller; accepting T&& may remove ownership from the caller (if a move ctor exists, etc.).
The auto && syntax uses two new features of C++11:
The auto part lets the compiler deduce the type based on the context (the return value in this case). This is without any reference qualifications (allowing you to specify whether you want T, T & or T && for a deduced type T).
The && is the new move semantics. A type supporting move semantics implements a constructor T(T && other) that optimally moves the content in the new type. This allows an object to swap the internal representation instead of performing a deep copy.
This allows you to have something like:
std::vector<std::string> foo();
So:
auto var = foo();
will perform a copy of the returned vector (expensive), but:
auto &&var = foo();
will swap the internal representation of the vector (the vector from foo and the empty vector from var), so will be faster.
This is used in the new for-loop syntax:
for (auto &item : foo())
std::cout << item << std::endl;
Where the for-loop is holding an auto && to the return value from foo and item is a reference to each value in foo.