We have Myclass &funcref(); and Myclass func(); and assign the return value to a variable (not a variable reference). If I understand correctly we use the copy constructor in both cases. The only difference is that funcref() returns a reference, and we copy over what is refers to into our object, while in func(), we copy the object while returning.
If we assign it to a variable reference, only func() still needs a copy constructor.
I hope everything until now is correct - if not, please point out my missunderstanding.
Now my question:
It seems like whether we assign the return value of func() to a variable, or a variable reference, it does not change anything at all? Is there any case in which those two will differ? At least when they are assigned, it seems like the process is the same, but I may be wrong.
In code:
class My_class {
//Anything may be here
};
My_class func();
My_class &funcref();
int main() {
My_class a = func(); //Copy operator used?
const My_class &b = func(); //Why does this need to be constant? Also, this uses copy operator, right?
My_class c = funcref(); //Copy operator used?
My_class &d = funcref(); //Here, we do not need a copy operator, right?
}
Main question: Except for b being const, is there any difference between a and b? Both seem to be on the stack of main, so do they differ in any way?
If I understand correctly we use the copy constructor in both cases.
Not quite. In C++11, move constructor will be used in case of Myclass func() (and compiler is allowed to optimise the copy / move away).
Since C++17, there is neither move nor copy in case of Myclass func().
Is there any case in which those two will differ?
It is somewhat unclear what cases you are asking about. There is a differenceiIn the case where the function returns a reference.
if I am calling any function that returns a non-reference, should I assign the output as a reference, or not as a reference?
Typically, you should use an object variable because that is simpler and doesn't involve the cognitive load of having to understand reference bound temporary lifetime extension rules.
Only case I can think of where reference variable for prvalue makes sense is a template context where the called function might return a reference, or a non-reference object that acts as a reference wrapper. I believe that case is the reason why temporary lifetime extension exists.
Related
A have a few questions regarding copy elision and move.
Let's assume I have the following code:
class A {};
class B {
public:
A get1() const {
A local_var;
return local_var;
}
A get1bad() const {
A local_var;
return std::move(local_var);
}
A get2() const {
return member_var;
}
A get3() const {
return std::move(member_var);
}
private:
A member_var;
};
I read a lot of people saying to not do move on return.
From what I gather it's because with copy elision, on case get1, the compiler will not call constructor + move constructor but rather just one call to the default constructor, while case get1bad forces the compiler to call constructor + move.
My question is regarding cases where the variable is not local (get2 vs get3).
In that case, the variable is constructed anyway in the class. In get2 there's not really any optimization that I can see being possible. In this case, if I don't really care about ensuring class B has a valid A object, wouldn't it be better to actually move? Wouldn't it just call the move constructor which is generally cheaper than the copy constructor?
A get3() const {
return std::move(member_var);
}
You have a const function, so the member is also considered const and will not move anyway.
You could consider A get3() && to move the member when you have a temporary B. Maybe, if that happens often.
Having a getter that always destroys the class' value, even when it is an lvalue, seems like a rare use case.
I read a lot of people saying to not do move on return. From what I gather it's because with copy elision, on case get1, the compiler will not call constructor + move constructor but rather just one call to the default constructor, while case get1bad forces the compiler to call constructor + move.
If the operand to a return statement is just the name of a local variable, then std::move is basically implicitly applied. So at best return std::move(local_var); is redundant. This is a special rule for return. It doesn't apply elsewhere.
However, using return std::move(local_var); makes it so that the mandatory copy elision rules do not apply anymore, which is why it is not only redundant, but even worse than just return local_var;.
My question is regarding cases where the variable is not local (get2 vs get3). In that case, the variable is constructed anyway in the class. In get2 there's not really any optimization that I can see being possible. In this case, if I don't really care about ensuring class B has a valid A object, wouldn't it be better to actually move? Wouldn't it just call the move constructor which is generally cheaper than the copy constructor?
In your examples all member functions are const qualified. Therefore the type of std::move(member_var) will be const A&&. The move constructor takes a A&&, not const A&&. Therefore no move will happen and the copy constructor will be used for both get2 and get3's return values.
If you remove the const qualifiers, then yes return std::move(member_var); makes sense in that case and allows a move construction rather than copy construction. Copy/move elision is not possible in either case, since variable is not local to the function.
It is unlikely that you wouldn't care about the state of member_var after calling a member function, so it is unlikely that this will make sense in practice. If at all I would qualify the member function with &&, so that it is only used if the class instance is an rvalue, which is the only case I can think of where one doesn't care about the object state after the call to the member function.
So in c++ if you assign the return value of a function to a const reference then the lifetime of that return value will be the scope of that reference. E.g.
MyClass GetMyClass()
{
return MyClass("some constructor");
}
void OtherFunction()
{
const MyClass& myClass = GetMyClass(); // lifetime of return value is until the end
// of scope due to magic const reference
doStuff(myClass);
doMoreStuff(myClass);
}//myClass is destructed
So it seems that wherever you would normally assign the return value from a function to a const object you could instead assign to a const reference. Is there ever a case in a function where you would want to not use a reference in the assignment and instead use a object? Why would you ever want to write the line:
const MyClass myClass = GetMyClass();
Edit: my question has confused a couple people so I have added a definition of the GetMyClass function
Edit 2: please don't try and answer the question if you haven't read this:
http://herbsutter.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/
If the function returns an object (rather than a reference), making a copy in the calling function is necessary [although optimisation steps may be taken that means that the object is written directly into the resulting storage where the copy would end up, according to the "as-if" principle].
In the sample code const MyClass myClass = GetMyClass(); this "copy" object is named myclass, rather than a temporary object that exists, but isn't named (or visible unless you look at the machine-code). In other words, whether you declare a variable for it, or not, there will be a MyClass object inside the function calling GetMyClass - it's just a matter of whether you make it visible or not.
Edit2:
The const reference solution will appear similar (not identical, and this really just written to explain what I mean, you can't actually do this):
MyClass __noname__ = GetMyClass();
const MyClass &myclass = __noname__;
It's just that the compiler generates the __noname__ variable behind the scenes, without actually telling you about it.
By making a const MyClass myclass the object is made visible and it's clear what is going on (and that the GetMyClass is returning a COPY of an object, not a reference to some already existing object).
On the other hand, if GetMyClass does indeed return a reference, then it is certainly the correct thing to do.
IN some compilers, using a reference may even add an extra memory read when the object is being used, since the reference "is a pointer" [yes, I know, the standard doesn't say that, but please before complaining, do me a favour and show me a compiler that DOESN'T implement references as pointers with extra sugar to make them taste sweeter], so to use a reference, the compiler should read the reference value (the pointer to the object) and then read the value inside the object from that pointer. In the case of the non-reference, the object itself is "known" to the compiler as a direct object, not a reference, saving that extra read. Sure, most compilers will optimise such an extra reference away MOST of the time, but it can't always do that.
One reason would be that the reference may confuse other readers of your code. Not everybody is aware of the fact that the lifetime of the object is extended to the scope of the reference.
The semantics of:
MyClass const& var = GetMyClass();
and
MyClass const var = GetMyClass();
are very different. Generally speaking, you would only use the
first when the function itself returns a reference (and is
required to return a reference by its very semantics). And you
know that you need to pay attention to the lifetime of the
object (which is not under your control). You use the second
when you want to own (a copy of) the object. Using the second
in this case is misleading, can lead to surprises (if the
function also returns a reference to an object which is
destructed earlier) and is probably slightly less efficient
(although in practice, I would expect both to generate exactly
the same code if GetMYClass returns by value).
Performance
As most current compilers elide copies (and moves), both version should have about the same efficiency:
const MyClass& rMyClass = GetMyClass();
const MyClass oMyClass = GetMyClass();
In the second case, either a copy or move is required semantically, but it can be elided per [class.copy]/31. A slight difference is that the first one works for non-copyable non-movable types.
It has been pointed out by Mats Petersson and James Kanze that accessing the reference might be slower for some compilers.
Lifetime
References should be valid during their entire scope just like objects with automatic storage are. This "should" of course is meant to be enforced by the programmer. So for the reader IMO there's no differences in the lifetimes implied by them. Although, if there was a bug, I'd probably look for dangling references (not trusting the original code / the lifetime claim for the reference).
In the case GetMyClass could ever be changed (reasonably) to return a reference, you'd have to make sure the lifetime of that object is sufficient, e.g.
SomeClass* p = /* ... */;
void some_function(const MyClass& a)
{
/* much code with many side-effects */
delete p;
a.do_something(); // oops!
}
const MyClass& r = p->get_reference();
some_function(r);
Ownership
A variable directly naming an object like const MyClass oMyClass; clearly states I own this object. Consider mutable members: if you change them later, it's not immediately clear to the reader that's ok (for all changes) if it has been declared as a reference.
Additionally, for a reference, it's not obvious that the object its referring to does not change. A const reference only implies that you won't change the object, not that nobody will change the object(*). A programmer would have to know that this reference is the only way of referring to that object, by looking up the definition of that variable.
(*) Disclaimer: try to avoid unapparent side effects
I don't understand what you want to achieve. The reason that T const& can be bound (on the stack) to a T (by value) which is returned from a function is to make it possible other function can take this temporary as an T const& argument. This prevents you from requirement to create overloads. But the returned value has to be constructed anyway.
But today (with C++11) you can use const auto myClass = GetMyClass();.
Edit:
As an excample of what can happen I will present something:
MyClass version_a();
MyClass const& version_b();
const MyClass var1 =version_a();
const MyClass var2 =version_b();
const MyClass var3&=version_a();
const MyClass var4&=version_b();
const auto var5 =version_a();
const auto var6 =version_b();
var1 is initialised with the result of version_a()
var2 is initialised with a copy of the object to which the reference returned by version_b() belongs
var3 holds a const reference to to the temoprary which is returned and extends its lifetime
var4 is initialised with the reference returned from version_b()
var5 same as var1
var6 same as var4
They are semanticall all different. var3 works for the reason I gave above. Only var5 and var6 store automatically what is returned.
there is a major implication regarding the destructor actually being called. Check Gotw88, Q3 and A3. I put everything in a small test program (Visual-C++, so forgive the stdafx.h)
// Gotw88.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
class A
{
protected:
bool m_destroyed;
public:
A() : m_destroyed(false) {}
~A()
{
if (!m_destroyed)
{
std::cout<<"A destroyed"<<std::endl;
m_destroyed=true;
}
}
};
class B : public A
{
public:
~B()
{
if (!m_destroyed)
{
std::cout<<"B destroyed"<<std::endl;
m_destroyed=true;
}
}
};
B CreateB()
{
return B();
}
int _tmain(int argc, _TCHAR* argv[])
{
std::cout<<"Reference"<<std::endl;
{
const A& tmpRef = CreateB();
}
std::cout<<"Value"<<std::endl;
{
A tmpVal = CreateB();
}
return 0;
}
The output of this little program is the following:
Reference
B destroyed
Value
B destroyed
A destroyed
Here a small explanation for the setup. B is derived from A, but both have no virtual destructor (I know this is a WTF, but here it's important). CreateB() returns B by value. Main now calls CreateB and first stores the result of this call in a const reference of type A. Then CreateB is called and the result is stored in a value of type A.
The result is interesting. First - if you store by reference, the correct destructor is called (B), if you store by value, the wrong one is called. Second - if you store in a reference, the destructor is called only once, this means there is only one object. By value results in 2 calls (to different destructors), which means there are 2 objects.
My advice - use the const reference. At least on Visual C++ it results in less copying. If you are unsure about your compiler, use and adapt this test program to check the compiler. How to adapt? Add copy / move constructor and copy-assignment operator.
I quickly added copy & assignment operators for class A & B
A(const A& rhs)
{
std::cout<<"A copy constructed"<<std::endl;
}
A& operator=(const A& rhs)
{
std::cout<<"A copy assigned"<<std::endl;
}
(same for B, just replace every capital A with B)
this results in the following output:
Reference
A constructed
B constructed
B destroyed
Value
A constructed
B constructed
A copy constructed
B destroyed
A destroyed
This confirms the results from above (please note, the A constructed results from B being constructed as B is derived from A and thus As constructor is called whenever Bs constructor is called).
Additional tests: Visual C++ accepts also the non-const reference with the same result (in this example) as the const reference. Additionally, if you use auto as type, the correct destructor is called (of course) and the return value optimization kicks in and in the end it's the same result as the const reference (but of course, auto has type B and not A).
i m trying to make a simple program ( & yes , it is a homework ) that can generate Dates , & like most of normal people : i made my Class attributes private , i tried to send the same type that i m working on to the constructor but the complier have not accept it , i did some research & i found out that in cases like that people generously send a const "type" reference to the constructor witch meant to me that have not understand OOP well
so why do we have to send the const " type " reference instead of just the types name to the constructor ? & please give me some links or websites for beginners
here is a peace of my Code :
class Date {
int d ;
int m ;
int y ;
public :
Date();
Date(int , int , int);
Date(const Date &);// my question is : why do we have to write this instead of Date( Date )
};
PS : sorry for my English
To paraphrase our question:
why do we have to write Date(const Date &) instead of Date(Date)?
I'm going to split this into two parts, the first answering why a copy constructor needs to take its argument per reference, the second why this needs to be a const reference.
The reason a copy constructor needs to take its argument per reference is that, for a function that's taking an argument per copy void f(T arg), when you call it f(obj), obj is copied into arg using T's copy constructor. So if you want to implement the copy constructor, you'd better not take the argument by copy, because this would call the copy constructor while invoking it, leading to an endless recursion. You can easily try this yourself:
struct tester {
tester(tester) {std::cout << "inside of erroneous copy ctor\n";}
};
int main()
{
tester t1;
std::cout << "about to call erroneous copy ctor\n";
tester t2(t1);
std::cout << "done with call erroneous copy ctor\n";
return 0;
}
That program should only ever write one line and then blow the stack.
Note: As Dennis points out in his comment, actually this program is not guaranteed to compile, so, depending on your compiler, you might not really be able to try it.
Bottom line: A copy constructor should take its argument by reference, because taking it per copy would require the copy constructor.
That leaves the question of why it is const T& and not simply T&? In fact, there are two reasons for that.
The logical reason is that, when you invoke the copy constructor, you do not expect the object copied from to change. In C++, if you want to express that something is immutable, you use const. This tells users that they can safely pass their precious objects to your copy constructor, because it won't do anything with it except read from it. As a nice side effect, if you implement the copy constructor and accidentally try to write to the object, the compiler throws an error message at you, reminding you of the promise made to the caller.
The other reason is that you cannot bind temporary objects to non-const references, you can only bind them to const references. A temporary object is, for example, what a function might return:
struct tester {
tester(tester& rhs) {std::cout << "inside of erroneous copy ctor\n";}
};
tester void f()
{
tester t;
return t;
}
When f() is called, a tester object is created inside, and a copy of it is then returned to the caller, which might then put it into another copy:
tester my_t = f(); // won't compile
The problem is that f() returns a temporary object, and in order to call the copy constructor, this temporary would need to bind to the rhs argument of tester's copy constructor, which is a non-const reference. But you cannot bind a temporary object to a non-const reference, so that code won't compile.
While you can work around this if you want (just don't copy the temporary, but bind it to a const reference instead, which extends the temporary's lifetime to the end of the reference's lifetime: const tester& my_t = f()), people expect to be able to copy temporaries of your type.
Bottom line: A copy constructor should take its argument by const reference, because otherwise users might not be willing or able to use it.
Here's one more fact: In the next C++ standard, you can overload functions for temporary objects, so-called rvalues. So you can have a special copy constructor that takes temporary objects overloading the "normal" copy constructor. If you have a compiler that already supports this new feature, you can try it out:
struct tester {
tester(const tester& rhs) { std::cout << "common copy ctor\n"; }
tester( tester&& rhs) { std::cout << "copy ctor for rvalues\n"; }
};
When you use the above code to invoke our f()
tester my_t = f();
the new copy constructor for rvalues should be called when the temporary object returned by the call to f() is copied to my_t and the regular copy constructor might be called in order to copy the t object from inside of f() to the returned temporary. (Note: you might have to disable your compiler's optimization in order to see this, as the compiler is allowed to optimize away all the copying.)
So what can you with this? Well, when you copy an rvalue, you know that the object copied from is going to be destroyed after the call to the copy constructor, so the copy constructor taking an rvalue (T&&) could just steal the values from the argument instead of copying them. Since the object is going to be destroyed anyway, nobody is going to notice.
For some classes (for example, for string classes), moving the value from one object to another could be much cheaper than copying them.
if I understood your question correctly, to avoid making copies/calling constructor of object.
void function(const T&); // does not create new T
void function(T&); // does not create newT, however T must be modifiable (lvalue)
void function(T); // creates new T
for simple types creating new copy is trivial (and often optimized away by compiler).
For complex object, creating new copy may be very expensive.
Hence you pass it by reference.
https://isocpp.org/wiki/faq/references
https://isocpp.org/wiki/faq/ctors
if you are asking why can not do the following:
struct type {
type(type);
};
Is because this would lead to infinite recursion, since constructor depends on itself
you can do this however
struct type {
type(type, int);
};
since this constructor is different from synthesized type(const type&)
http://en.wikipedia.org/wiki/Copy_constructor
In addition to #aaa's answer, I will try to answer the const part. The const part simply means that the object you are passing logically does not change. This makes sense, because when a copy constructor is called with a Date object argument d, d should not be modified at all!
You can remove the const and your code will still work the same way. However, const provides the additional security that you can never modify the variable marked as const. In your case, this means you can not call any of the non-const method of Date. This is enforced by the compiler at compile-time.
Historically this is the reason for introducing references to the language. Here's an explanation:
In C you can pass values to parameters by value (void f(struct custom_type i)) or by pointer (void g(struct custom_type* i)).
With POD values (int, char, etc.) passing by value is not a problem, but if you are looking at complex structures, then the stack grows too quickly by placing entire structures on stack for function calls. That is why in C you tend to pass structures as parameters by pointer, even if the function doesn't modify them.
In C++ there are cases where neither option worked:
passing by pointers involves a counter-intuitive syntax for operators (if you define operator + for a class custom_type writing custom_type a, b, c; a = &b + &c; is counterintuitive as a doesn't get assigned the sum of the addresses. Furthermore if you wanted to be able to assign the sum of the values to a and the sum of the addresses to a, you would have to somehow differentiate between the cases, by syntax).
passing by value is impossible or undesired in the case of copy constructors. In your case, if you have Date(Date d) {} and assignment Date a; Date b(a); what you get is that a copy of a is created just to be passed as a parameter to the constructor of b. This leads to infinite recursion, as creating a copy of a to pass as a parameter involves is the same as Date d = a; b = Date(d);.
For these reasons (and there may have been others) a decision was made to create references: data types that looks syntactically like a value type, but behave like pointers (that is, it points to the value of another variable, but you access it like a variable, not like a pointer).
Regarding the reason why you need const in the declaration, it is so that your constructor will accept temporary objects. As you cannot modify the value of a temporary references if your constructor doesn't accept const& you can only use the copy constructor for non-const stable objects.
That is, if you have:
class Date
{
public:
Date(Date& other); // non-const reference
...
you can write:
Date a;
Date b = a;
but not:
Date someFunction() { return Date(xxx); }
Date a = someFunction(); // someFunction returns a temporary object
neither:
const Date someImportantDate;
Date a = someImportantDate; // cannot pass const value to non-const
Is it valid to store the return value of an object in a reference?
class A { ... };
A myFunction()
{
A myObject;
return myObject;
} //myObject goes out of scope here
void mySecondFunction()
{
A& mySecondObject = myFunction();
}
Is it possible to do this in order to avoid copying myObject to mySecondObject? myObject is not needed anymore and should be exactly the same as mySecondObject so it would in theory be faster just to pass ownership of the object from one object to another. (This is also possible using boost shared pointer but that has the overhead of the shared pointer.)
Thanks in advance.
It is not allowed to bind the temporary to a non-const reference, but if you make your reference const you will extend the lifetime of the temporary to the reference, see this Danny Kalev post about it.
In short:
const A& mySecondObject = myFunction();
It's possible with a const reference.
myFunction returns by value, so that return value is a temporary object. You can bind a temporary to a const reference, and the lifetime of the temporary is extended to the lifetime of the reference. You can't bind a temporary to a non-const reference (unfortunately).
The return value of myFunction might be a copy of myObject. On the plus side, copy constructor elision (aka the "named return value optimisation" in this case) permits the compiler to construct myObject directly into the temporary which is the return value of myFunction, presumably located somewhere on the stack of the calling code. If it does that, then when myObject goes out of scope the object is not actually destroyed. The optimisation is commonly implemented - for example GCC (usually?) does it even without any optimisation flags.
Copy ctor elision also permits the compiler to avoid all copying if you did:
A mySecondObject = myFunction();
This requires the application of both legal types of copy ctor elision: (1) returning a named value from a function, and (2) initializing an object from a temporary.
You might be interested in the return-by-value optimization that many compilers make to avoid calling the copy constructor.
Is there a difference between foo and bar:
class A
{
Object __o;
void foo(Object& o)
{
__o = o;
}
void bar(Object o)
{
__o = o;
}
}
As I understand it, foo performs no copy operation on object o when it is called, and one copy operation for assignment. Bar performs one copy operation on object o when it is called and another one for assignment. So I can more or less say that foo uses 2 times less memory than bar (if o is big enough). Is that correct ?
Is it possible that the compiler optimises the bar function to perform only one copy operation on o ? i.e. makes __o pointing on the local copy of argument o instead of creating a new copy?
It depends. For example, if the compiler decides to inline the function, obviously there will be no copy since there is no function call.
If you want to be sure, pass by const-reference:
void bar(const Object& o)
This makes no copies. Note your non-const version requires an lvalue, because the reference is mutable. foo(Object()) wouldn't work, but temporaries (rvalues) can be bound to a const-reference.
Double-underscores in identifiers are reserved for the compiler, by the way.
Since assignment typically takes const Something& as a parameter, it would be canonical to write:
void foo(const Object& o)
{
__o = o;
}
But this does not answer your question about the optimization. I'm not sure that, in general, such optimization can / will be made by the compiler.
I think its fair to say that foo() performs one less copy than bar(). Its not very meaningful to say how much more or less memory is consumed, because for simple objects they are stored on the stack and cleaned up after return from bar().
As has been said, don't use underscores at the start of identifiers.
I think compilers can do the optimization in case of temporary objects. This technique is called copy elision.
Please refer to the answers to question i had posted what-is-copy-elision-and-how-it-optimizes-copy-and-swap-idiom. This ans is really helpful What is copy elision and how does it optimize the copy-and-swap idiom?
Though i am not an expert in this, from what i understand compliers can optimize copying of temporary object in some scenarios.
For example if your code is called like this
bar(getObject())
where getObject has a signature
Object getObject()
This call will result in creation of a temporary of type Object. If the compiler does not do any optimization that temporay has to be copied into the argument to bar.
However if compiler supports copy elision this copying wont be performed and the temporary will be passed as an argument to bar function. So the copy is avoided and its performance is same as foo which accepts a reference. But as i said it happens only in case of temporary object