I was reading a book about data structure implemented in C++, I dont understand a code snippet， it's part of vector class
void push_back(object &&x) {
//do something
objects[size++] = std::move(x);
}
I know that std::move return a rvalue reference of the object, but the push_back member function already has rvalue reference x as parameter, isn't the std::move here unnecessary？
Another question is if we have a rvalue reference of a class object, we still need to use std::move on its member if we want to call move instead of copy right? like the code below:
A& operator=(A&& other) {
member = std::move(other.member);
return *this;
}
isn't the std::move here unnecessary？
No. Types and value categories are different things.
(emphasis mine)
Each C++ expression (an operator with its operands, a literal, a variable name, etc.) is characterized by two independent properties: a type and a value category.
The following expressions are lvalue expressions:
the name of a variable, a function, a template parameter object (since
C++20), or a data member, regardless of type, such as std::cin or
std::endl. Even if the variable's type is rvalue reference, the
expression consisting of its name is an lvalue expression;
std::move converts lvalue to rvalue (xvalue). As a named variable, x is an lvalue, std::move converts it to rvalue in objects[size++] = std::move(x); then the move assignment operator is supposed to be used. Otherwise, copy assignment operator will be used instead; lvalue can't be bound to rvalue reference.
we still need to use std::move on its member if we want to call move instead of copy right?
Yes, same reason as above.
x has a name, thus it's an lvalue inside the function. The rvalue reference was bound to the lvalue x. std::move casts it back to the rvalue that was passed in.
Related
I'm using the embedded templated library with etl::queue
https://www.etlcpp.com/queue.html
the etl::queue is quitvalent to std::queue
In order to avoid copying I'd like to actually move the element into the queue.
Now my setup looks like this
bool CETLConcurrentQueue<ElementType_T, u32QueueSize>::Add(ElementType_T &&element, const uint32_t u32Timeout)
{
//lock mutex...
queue.push(element);
//do further stuff
}
Now I'm not using queue.push(std::move(element)); because element is allready an rvalue Reference
However, queue.push(element); calls elements copy constructor (which is deleted)
How can I call elements move constructor instead?
You have to use std::move to convert element to rvalue. As a named variable element itself is an lvalue, even its type is an rvalue reference.
queue.push(std::move(element));
Note that types and value categories are two independent things.
(emphasis mine)
Each C++ expression (an operator with its operands, a literal, a variable name, etc.) is characterized by two independent properties: a type and a value category.
...
The following expressions are lvalue expressions:
the name of a variable, a function, a template parameter object (since C++20), or a data member, regardless of type, such as std::cin
or std::endl. Even if the variable's type is rvalue reference, the
expression consisting of its name is an lvalue expression;
I think there's something I'm not quite understanding about rvalue references. Why does the following fail to compile (VS2012) with the error 'foo' : cannot convert parameter 1 from 'int' to 'int &&'?
void foo(int &&) {}
void bar(int &&x) { foo(x); };
I would have assumed that the type int && would be preserved when passed from bar into foo. Why does it get transformed into int once inside the function body?
I know the answer is to use std::forward:
void bar(int &&x) { foo(std::forward<int>(x)); }
so maybe I just don't have a clear grasp on why. (Also, why not std::move?)
I always remember lvalue as a value that has a name or can be addressed. Since x has a name, it is passed as an lvalue. The purpose of reference to rvalue is to allow the function to completely clobber value in any way it sees fit. If we pass x by reference as in your example, then we have no way of knowing if is safe to do this:
void foo(int &&) {}
void bar(int &&x) {
foo(x);
x.DoSomething(); // what could x be?
};
Doing foo(std::move(x)); is explicitly telling the compiler that you are done with x and no longer need it. Without that move, bad things could happen to existing code. The std::move is a safeguard.
std::forward is used for perfect forwarding in templates.
Why does it get transformed into int once inside the function body?
It doesn't; it's still a reference to an rvalue.
When a name appears in an expression, it's an lvalue - even if it happens to be a reference to an rvalue. It can be converted into an rvalue if the expression requires that (i.e. if its value is needed); but it can't be bound to an rvalue reference.
So as you say, in order to bind it to another rvalue reference, you have to explicitly convert it to an unnamed rvalue. std::forward and std::move are convenient ways to do that.
Also, why not std::move?
Why not indeed? That would make more sense than std::forward, which is intended for templates that don't know whether the argument is a reference.
It's the "no name rule". Inside bar, x has a name ... x. So it's now an lvalue. Passing something to a function as an rvalue reference doesn't make it an rvalue inside the function.
If you don't see why it must be this way, ask yourself -- what is x after foo returns? (Remember, foo is free to move x.)
rvalue and lvalue are categories of expressions.
rvalue reference and lvalue reference are categories of references.
Inside a declaration, T x&& = <initializer expression>, the variable x has type T&&, and it can be bound to an expression (the ) which is an rvalue expression. Thus, T&& has been named rvalue reference type, because it refers to an rvalue expression.
Inside a declaration, T x& = <initializer expression>, the variable x has type T&, and it can be bound to an expression (the ) which is an lvalue expression (++). Thus, T& has been named lvalue reference type, because it can refer to an lvalue expression.
It is important then, in C++, to make a difference between the naming of an entity, that appears inside a declaration, and when this name appears inside an expression.
When a name appears inside an expression as in foo(x), the name x alone is an expression, called an id-expression. By definition, and id-expression is always an lvalue expression and an lvalue expressions can not be bound to an rvalue reference.
When talking about rvalue references it's important to distinguish between two key unrelated steps in the lifetime of a reference - binding and value semantics.
Binding here refers to the exact way a value is matched to the parameter type when calling a function.
For example, if you have the function overloads:
void foo(int a) {}
void foo(int&& a) {}
Then when calling foo(x), the act of selecting the proper overload involves binding the value x to the parameter of foo.
rvalue references are only about binding semantics.
Inside the bodies of both foo functions the variable a acts as a regular lvalue. That is, if we rewrite the second function like this:
void foo(int&& a) {
foo(a);
}
then intuitively this should result in a stack overflow. But it doesn't - rvalue references are all about binding and never about value semantics. Since a is a regular lvalue inside the function body, then the first overload foo(int) will be called at that point and no stack overflow occurs. A stack overflow would only occur if we explicitly change the value type of a, e.g. by using std::move:
void foo(int&& a) {
foo(std::move(a));
}
At this point a stack overflow will occur because of the changed value semantics.
This is in my opinion the most confusing feature of rvalue references - that the type works differently during and after binding. It's an rvalue reference when binding but it acts like an lvalue reference after that. In all respects a variable of type rvalue reference acts like a variable of type lvalue reference after binding is done.
The only difference between an lvalue and an rvalue reference comes when binding - if there is both an lvalue and rvalue overload available, then temporary objects (or rather xvalues - eXpiring values) will be preferentially bound to rvalue references:
void goo(const int& x) {}
void goo(int&& x) {}
goo(5); // this will call goo(int&&) because 5 is an xvalue
That's the only difference. Technically there is nothing stopping you from using rvalue references like lvalue references, other than convention:
void doit(int&& x) {
x = 123;
}
int a;
doit(std::move(a));
std::cout << a; // totally valid, prints 123, but please don't do that
And the keyword here is "convention". Since rvalue references preferentially bind to temporary objects, then it's reasonable to assume that you can gut the temporary object, i.e. move away all of its data away from it, because after the call it's not accessible in any way and is going to be destroyed anyway:
std::vector<std::string> strings;
string.push_back(std::string("abc"));
In the above snippet the temporary object std::string("abc") cannot be used in any way after the statement in which it appears, because it's not bound to any variable. Therefore push_back is allowed to move away its contents instead of copying it and therefore save an extra allocation and deallocation.
That is, unless you use std::move:
std::vector<std::string> strings;
std::string mystr("abc");
string.push_back(std::move(mystr));
Now the object mystr is still accessible after the call to push_back, but push_back doesn't know this - it's still assuming that it's allowed to gut the object, because it's passed in as an rvalue reference. This is why the behavior of std::move() is one of convention and also why std::move() by itself doesn't actually do anything - in particular it doesn't do any movement. It just marks its argument as "ready to get gutted".
The final point is: rvalue references are only useful when used in tandem with lvalue references. There is no case where an rvalue argument is useful by itself (exaggerating here).
Say you have a function accepting a string:
void foo(std::string);
If the function is going to simply inspect the string and not make a copy of it, then use const&:
void foo(const std::string&);
This always avoids a copy when calling the function.
If the function is going to modify or store a copy of the string, then use pass-by-value:
void foo(std::string s);
In this case you'll receive a copy if the caller passes an lvalue and temporary objects will be constructed in-place, avoiding a copy. Then use std::move(s) if you want to store the value of s, e.g. in a member variable. Note that this will work efficiently even if the caller passes an rvalue reference, that is foo(std::move(mystring)); because std::string provides a move constructor.
Using an rvalue here is a poor choice:
void foo(std::string&&)
because it places the burden of preparing the object on the caller. In particular if the caller wants to pass a copy of a string to this function, they have to do that explicitly;
std::string s;
foo(s); // XXX: doesn't compile
foo(std::string(s)); // have to create copy manually
And if you want to pass a mutable reference to a variable, just use a regular lvalue reference:
void foo(std::string&);
Using rvalue references in this case is technically possible, but semantically improper and totally confusing.
The only, only place where an rvalue reference makes sense is in a move constructor or move assignment operator. In any other situation pass-by-value or lvalue references are usually the right choice and avoid a lot of confusion.
Note: do not confuse rvalue references with forwarding references that look exactly the same but work totally differently, as in:
template <class T>
void foo(T&& t) {
}
In the above example t looks like a rvalue reference parameter, but is actually a forwarding reference (because of the template type), which is an entirely different can of worms.
See the code snippet below:
#include<vector>
using std::vector;
int main()
{
vector<int> a;
auto &&it=--a.end();//what?the hint tell me it deduces a lvalue reference type!
}
We all know that a self-defined class with operator -- returns a xvalue when the operand itself is a rvalue, like --Myclass(). Obviously Myclass() is prvalue, so the return value of --Myclass() should also be rvalue(prcisely,xvalue) ,too.
from cppref
a.m, the member of object expression, where a is an rvalue and m is a non-static data member of non-reference type;
So why does the auto deduce lvalue reference in this circumstance?
What's more,the code snippet could be compiled without any error!
Why could a rvalue be binded to a lvalue reference?
And I've come across a confusing error(Not the same as the code snippet above,I'm sure the vector is not empty):Later when I use the it, segment fault happens!
The code that causes segment fault (in the last three lines) This code is the answer for a Chinese online testing PAT,when I submit the answer, it arouses segment fault.
Is it undefined behavior to use lvalue reference to bind a --Myclass() ,and use it later?
We all know that a self-defined class with operator -- returns a rvalue when we write somethin like --Myclass()
No, that's not something we know. Technically, a user defined prefix decrement operator can return either an object, or a reference. In practice, the prefix decrement operator typically returns an lvalue reference.
Obviously Myclass() is prvalue, so the return value of --Myclass() should also be prvalue ,too.
That's not how value categories propagate. You can call a function on a prvalue, and the function can return an lvalue (possibly to *this).
So why does the auto deduce lvalue reference in this circumstance?
Because the decrement operator of the iterator returns an lvalue reference.
Use auto it to fix the problem.
Is it undefined behavior to use lvalue reference to bind a --Myclass() ,and use it later?
Depends on how Myclass::operator--() is declared. If it returns an lvalue reference, then that is UB. If it returns an object, then there is no UB. It is possible to provide both variants overloaded by ref-qualifier.
This is an "universal reference". The variable is not an rvalue reference, but an lvalue into which the rvalue is moved. This is necessary because a reference would not extend the life time of the temporary object.
I was wondering about a c++ behaviour when an r-value is passed among functions.
Look at this simple code:
#include <string>
void foo(std::string&& str) {
// Accept a rvalue of str
}
void bar(std::string&& str) {
// foo(str); // Does not compile. Compiler says cannot bind lvalue into rvalue.
foo(std::move(str)); // It feels like a re-casting into a r-value?
}
int main(int argc, char *argv[]) {
bar(std::string("c++_rvalue"));
return 0;
}
I know when I'm inside bar function I need to use move function in order to invoke foo function. My question now is why?
When I'm inside the bar function the variable str should already be an r-value, but the compiler acts like it is a l-value.
Can somebody quote some reference to the standard about this behaviour?
Thanks!
str is a rvalue reference, i.e. it is a reference only to rvalues. But it is still a reference, which is a lvalue. You can use str as a variable, which also implies that it is an lvalue, not a temporary rvalue.
An lvalue is, according to §3.10.1.1:
An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment expression) designates a function or an object. [ Example: If E is an expression of pointer type, then *E is an lvalue expression referring to the object or function to which E points. As another example, the result of calling a function whose return type is an lvalue reference is an lvalue. —end example ]
And an rvalue is, according to §3.10.1.4:
An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment
expression) is an xvalue, a temporary object (12.2) or subobject thereof, or a value that is not associated with an object.
Based on this, str is not a temporary object, and it is associated with an object (with the object called str), and so it is not an rvalue.
The example for the lvalue uses a pointer, but it is the same thing for references, and naturally for rvalue references (which are only a special type of references).
So, in your example, str is an lvalue, so you have to std::move it to call foo (which only accepts rvalues, not lvalues).
The "rvalue" in "rvalue reference" refers to the kind of value that the reference can bind to:
lvalue references can bind to lvalues
rvalue references can bind to rvalues
(+ a bit more)
That's all there's to it. Importantly, it does not refer to the value that get when you use the reference. Once you have a reference variable (any kind of reference!), the id-expression naming that variable is always an lvalue. Rvalues occur in the wild only as either temporary values, or as the values of function call expressions, or as the value of a cast expression, or as the result of decay or of this.
There's a certain analogy here with dereferencing a pointer: dereferencing a pointer is always an lvalue, no matter how that pointer was obtained: *p, *(p + 1), *f() are all lvalues. It doesn't matter how you came by the thing; once you have it, it's an lvalue.
Stepping back a bit, maybe the most interesting aspect of all this is that rvalue references are a mechanism to convert an rvalue into an lvalue. No such mechanism had existed prior to C++11 that produced mutable lvalues. While lvalue-to-rvalue conversion has been part of the language since its very beginnings, it took much longer to discover the need for rvalue-to-lvalue conversion.
My question now is why?
I'm adding another answer because I want to emphasize an answer to the "why".
Even though named rvalue references can bind to an rvalue, they are treated as lvalues when used. For example:
struct A {};
void h(const A&);
void h(A&&);
void g(const A&);
void g(A&&);
void f(A&& a)
{
g(a); // calls g(const A&)
h(a); // calls h(const A&)
}
Although an rvalue can bind to the a parameter of f(), once bound, a is now treated as an lvalue. In particular, calls to the overloaded functions g() and h() resolve to the const A& (lvalue) overloads. Treating a as an rvalue within f would lead to error prone code: First the "move version" of g() would be called, which would likely pilfer a, and then the pilfered a would be sent to the move overload of h().
Reference.
I was wondering if there is a point in having a rvalue reference variable (not as a function parameter)? I understand the use of rvalue reference when it is used as a function variable as then it is possible to avoid unnecessary allocations etc. But is there a use case rvalue reference variables that are not function parameters? Initially I thought that with such variables we could capture data that was passed in as rvalue for use later, but it seems that if we have a rvalue reference variable, then we can already take its address and therefore it can not be rvalue reference anymore.
I tried the following code and it ( unsurprisingly ) does not compile. So why would I ever want to have a rvalue reference variable?
void test(int&& a) {}
int&& a(22);
test(a);
Thanks!
Your a here is the name of a rvalue reference. a itself is a lvalue (you can take its address) of type rvalue reference (yes, a bit confusing). So whenever you pass your a to test(a), you pass a lvalue. Think about lvalues as any object that has a name, or of which you can latter take its address. And you cannot bind a lvalue to a rvalue reference, hence the compile error
error: cannot bind 'int' lvalue to 'int&&'
You need to pass std::move(a). In this case, you cast away the lvalue-ness of a.
Note that inside your function a is also a lvalue (again, you can perform lvalue operations on it, like taking its address etc). You need a cast (std::move) anytime you want to use it as a rvalue. For example, say you have an object Bar with a member variable Foo member;, and you want to move into it via a member function f:
void Bar::f(Foo&& param)
{
member = std::move(param); // need std::move here to move param into member
}
// invoked it as
bar.f(std::move(some_foo)); // move some_foo into param
If you don't cast param to a rvalue reference with std::move, then you don't move param into member, you just copy it (if param is copy-able, otherwise you get a compile-time error), since param itself is a lvalue. Hope this clarifies it.