const_cast a const member in a class constructor - c++

I sometimes use const_cast when I want a member variable of a class to be constant during the life of the class, but it needs to be mutable during the constructor. Example:
struct qqq {
const vector<foo> my_foo;
qqq(vector<foo>* other) {
vector<foo>& mutable_foo = const_cast<vector<foo>&>(my_foo)
other->swap(mutable_foo);
}
};
I had assumed that doing this in the constructor was basically OK because nobody else is relying on it at this point so it wouldn't interact badly with optimization, etc.
However recently someone told me this is "undefined behavior" and that it's basically illegal to mutate a const object after it's been constructed under any circumstance.
Can someone clarify? Is this a bad / undefined behavior / thing to do?

It is Undefined Behavior. Per Paragraph 7.1.6.1/4 of the C++11 Standard:
Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const
object during its lifetime (3.8) results in undefined behavior.
In this case, it seems like you want your object to "become" constant after construction. This is not possible.
If your vector is meant to be const, you shall initialize it in the constructor's initialization list:
qqq(vector<foo>& other)
: my_foo(std::move(other))
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^
{
}
Notice, that unless you have a good reason for passing by pointer - in which case, you should also be checking whether the pointer is non-null - you should consider passing by reference (as shown above), which is the common practice.
UPDATE:
As Pete Becker correctly points out in the comments, proper design would suggest that the decision to move from the vector argument should belong to the caller of qqq's constructor, and not to the constructor itself.
If the constructor is always supposed to move from its argument, then you could let it accept an rvalue reference, making it clear what the constructor itself is expecting out of the caller:
qqq(vector<foo>&& other)
// ^^
: my_foo(std::move(other))
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^
{
}
This way, the caller would have to provide an rvalue in input to qqq's constructor:
std::vector<foo> v;
// ...
qqq q1(v); // ERROR!
qqq q2(std::move(v)); // OK! Now the client is aware that v must be moved from

Yes, it's indeed UB (Undefined Behaviour). You cannot modify a const object once it's initialised. What you should do is use the member initialiser list, perhaps together with a function:
struct qqq {
const vector<foo> my_foo;
qqq(vector<foo> *other) : my_foo(initialiseFoo(*other)) {}
static vector<foo> initialiseFoo(vector<foo> &other) {
vector<foo> tmp;
other.swap(tmp);
return tmp;
}
};
A decent optimiser should be able to get rid of the temporary.
If you can use C++11, it's actually even simpler:
struct qqq {
const vector<foo> my_foo;
qqq(vector<foo> *other) : my_foo(std::move(*other))
{
other->clear(); //Just in case the implementation of moving vectors is really weird
}
};

Related

Binding const& of temporary: No compiler warning?

I have a TestClass with a const& member variable. I know from various places and own experiences that it is a bad idea to initialize this const& with the reference to a temporary value. So I was quite suprised that the following code will compile fine (tested with gcc-4.9.1, clang-3.5, and scan-build-3.5) but fail to run properly.
class TestClass {
public:
// removing the "reference" would remove the temporary-problem
const std::string &d;
TestClass(const std::string &d)
: d(d) {
// "d" is a const-ref, cannot be changed at all... if it is assigned some
// temporary value it is mangled up...
}
};
int main() {
// NOTE: the variable "d" is a
// temporary, whose reference is not valid... what I don't get in the
// moment: why does no compiler warn me?
TestClass dut("d");
// and printing what we got:
std::cout << "beginning output:\n\n";
// this will silently abort the program (gcc-4.9.1) or be empty
// (clang-3.5) -- don't know whats going on here...
std::cout << "dut.d: '" << dut.d << "'\n";
std::cout << "\nthats it!\n";
return 0;
}
Why does none of the two compilers warn me at compile-time? See also this ideone, with some more testing going on.
No warning as no offense:
local const references prolong the lifespan of the variable.
The standard specifies such behavior in §8.5.3/5, [dcl.init.ref], the section on initializers of reference declarations. The lifetime extension is not transitive through a function argument. §12.2/5 [class.temporary]:
The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is
the complete object to a subobject of which the temporary is bound
persists for the lifetime of the reference except as specified below.
A temporary bound to a reference member in a constructor’s
ctor-initializer (§12.6.2 [class.base.init]) persists until the
constructor exits. A temporary bound to a reference parameter in a
function call (§5.2.2 [expr.call]) persists until the completion of
the full expression containing the call.
You can have a look at gotw-88 for an extended and more readable discussion on this topic.
Undefined Behaviour
So is it your code correct? Nope, and its execution will lead to undefined behaviour. The real problem in your code snapshot is that the Undefined Behaviour is caused by the mix of two perfectly legal operations: the call of the constructor passing a temporary object (whose life spans inside the constructor block) and the binding of the reference in the constructor definition.
The compiler is not smart enough to detect this explosive statement combination, so this is why you don't get any warning.
Binding a const & to a temporary is valid and the compiler will ensure that the temporary will live at least as long as the reference. This allows you to do things like pass string literals into functions expecting a const std::string &.
In your case however you are copying that reference and thus the lifetime guarantee no longer holds. Your constructor exits and the temporary is destroyed and you are left with a reference to invalid memory.
The problem is that there is no single point in which a warning would be warranted. It's only the combination of the call of the constructor and its implementation that leads to Undefined Behaviour.
If you consider just the constructor:
class TestClass {
public:
const std::string &d;
TestClass(const std::string &d)
: d(d)
{}
};
There's nothing wrong here, you got a reference and you're storing one. Here's an example of perfectly valid use:
class Widget {
std::string data;
TestClass test;
public:
Widget() : data("widget"), test(data)
{}
};
If you consider just the call site:
//Declaration visible is:
TestClass(const std::string &d);
int main() {
TestClass dut("d");
}
Here, the compiler doesn't "see" (in the general case) the definition of the constructor. Imagine an alternative:
struct Gadget {
std::string d;
Gadget(cosnt std::string &d) : d(d) {}
};
int main()
{
Gadget g("d");
}
Surely you wouldn't want a warning here either.
To summarise, both the call site and the constructor implementation are perfectly usable as-is. It's only their combination which causes issues, but that combination is beyond the context a compiler can reasonably use to emit warnings.
TestClass(const std::string &d1)
: d(d1) {
TestClass dut("d");
I guess following is happening logically:-
1) Your string literal ("d") would be implicitly converted to std::string ( Let's give it a name 'x' ).
2) So, 'x' is a temporary which is bound to d1 here. Lifetime of this temporary is extended to lifetime of your d1. Although that string literal would always be alive till the end of program.
3) Now you're making 'd' refer to 'd1'.
4) At the end of your constructor d1's lifetime is over and so is d's.
All compiler's are not so clever to figure out these minor glitches...

why use a const non-reference when const reference lifetime is the length of the current scope

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).

Is it good practice to bind shared pointers returned by functions to lvalue references to const?

Although it took me a while to get used to it, I now grew the habit of letting my functions take shared pointer parameters by lvalue-reference to const rather than by value (unless I need to modify the original arguments, of course, in which case I take them by lvalue-reference to non-const):
void foo(std::shared_ptr<widget> const& pWidget)
// ^^^^^^
{
// work with pWidget...
}
This has the advantage of avoiding an unnecessary copy of a shared pointer, which would mean thread-safely increasing the reference counting and potentially incurring in undesired overhead.
Now I've been wondering whether it is sane to adopt a somewhat symmetrical habit for retrieving shared pointers that are returned by value from functions, like at the end of the following code snippet:
struct X
{
// ...
std::shared_ptr<Widget> bar() const
{
// ...
return pWidget;
}
// ...
std::shared_ptr<Widget> pWidget;
};
// ...
// X x;
std::share_ptr<Widget> const& pWidget = x.bar();
// ^^^^^^
Are there any pitfalls with adopting such a coding habit? Is there any reason why I should prefer, in general, assigning a returned shared pointer to another shared pointer object rather than binding it to a reference?
This is just a remake of the old question of whether capturing a const reference to a temporary is more efficient than creating a copy. The simple answer is that it isn't. In the line:
// std::shared_ptr<Widget> bar();
std::shared_ptr<Widget> const & pWidget = bar();
The compiler needs to create a local unnamed variable (not temporary), initailize that with the call to bar() and then bind the reference to it:
std::shared_ptr<Widget> __tmp = bar();
std::shared_ptr<Widget> const & pWidget = __tmp;
In most cases it will avoid the creation of the reference and just alias the original object in the rest of the function, but at the end of the day whether the variable is called pWidget or __tmp and aliased won't give any advantage.
On the contrary, for the casual reader, it might look like bar() does not create an object but yield a reference to an already existing std::shared_ptr<Widget>, so the maintainer will have to seek out where bar() is defined to understand whether pWidget can be changed outside of the scope of this function.
Lifetime extension through binding to a const reference is a weird feature in the language that has very little practical use (namely when the reference is of a base and you don't quite care what the exact derived type returned by value is, i.e. ScopedGuard).
You may have the optimization backwards:
struct X
{
// ...
std::shared_ptr<Widget> const& bar() const
{
// ...
return pWidget;
}
// ...
std::shared_ptr<Widget> pWidget;
};
// ...
// X x;
std::share_ptr<Widget> pWidget = x.bar();
As bar is returning a member variable, it must take a copy of the shared_ptr in your version. If you return the member variable by reference the copy can be avoided.
This doesn't matter in both your original version and the version shown above, but would come up if you called:
x.bar()->baz()
In your version a new shared_ptr would be created, and then baz would be called.
In my version baz is called directly on the member copy of the shared_ptr, and the atomic reference increment/decrement is avoided.
Of course the cost of the shared_ptr copy constructor (atomic increment) is very small, and not even noticable in all but the most performance-sensetive applications. If you are writing a performance sensetive application than the better option would be to manage memory manually with a memory pool architecture and then to (carefully) use raw pointers instead.
Adding on top of what David Rodríguez - dribeas said namely, binding to a const reference doesn't save you from making the copy and the counter is incremented anyway, the following code illustrates this point:
#include <memory>
#include <cassert>
struct X {
std::shared_ptr<int> p;
X() : p{new int} {}
std::shared_ptr<int> bar() { return p; }
};
int main() {
X x;
assert(x.p.use_count() == 1);
std::shared_ptr<int> const & p = x.bar();
assert(x.p.use_count() == 2);
return 0;
}

Is return by value always const?

This code does not compile:
class C {};
void foo (C& c) {}
C bar() { return C(); }
int main()
{
foo(bar());
}
Compilation error (GCC 4.1.2) in line foo(bar()):
invalid initialization of non-const reference of type 'C&'
from a temporary of type 'C'
As bar() returns a mutable object, it should compile...
Why C++ does not allow this above code?
EDIT: I have summarize in an answer below all good ideas from all answers ;-)
The applicable rule here is that you can't create a non-const reference to a temporary object. If foo was declared as foo(const C&) the code would be okay.
The temporary object itself is not const, though; you can call non-const member functions on it, e.g., bar().non_const_member_function().
With C++11, foo can be written to take an rvalue reference; in that case, the call would be okay:
void foo(C&&);
foo(bar()); // okay
It's because the value returned by bar is a temporary value. As it's existence is temporary, you can't use a pointer or reference to that.
However, if you store a copy of that temporary, as in your second change, you no longer pass a reference to a temporary object to foo, but a reference to a real tangible object. And in the first case, when you change to a reference to a constant object, the compiler makes sure the temporary object stays around long enough (as per the C++ specification).
The issue is not with the declaration of bar but with that of foo. foo takes a non-const reference, and temporaries can only bind to const references (which then extends the lifetime of the temporary to match that of the reference it is bound to).
Allowing a non-const reference to bind to a temporary doesn't make much sense. A non-const reference implies that it will modify whatever object is bound to it. Modifying a temporary serves no purpose since its lifetime is limited and the changes will be lost as soon as it goes out of scope.
Modifiable (lvalue-)references do not bind to temporary values. However, const-references do bind to temporary values. It has nothing to do with whether the object returned by value is const or not; it's simply a matter of whether the expression is temporary or not.
For example, the following is valid:
struct C { void i_am_non_const() {} };
int main()
{
bar().i_am_non_const();
}
It is a design choice. There is nothing inherently impossible here. Just a design choice.
In C++11, you have a third alternative which is also superior alternative:
void foo(C && c) {}
That is, use rvalue-references.
It's not const, but it is a temporary rvalue. As such, it can't bind to a non-const lvalue reference.
It can bind to a const or rvalue reference, and you can call member functions (const or not) on it:
class C { void f(); };
void foo_const(C const &);
void foo_rvalue(C &&);
foo_const( bar() ); // OK
foo_rvalue( bar() ); // OK
bar().f(); // OK
The real, hard truth is that it makes no sense to get a reference to a temporary value.
The big point of passing an object by reference is that it allows you to modify its state. However, in the case of a temporary, by its very nature, it would not be particularly helpful to be able to modify it, since you have no way of getting another reference to it later in your code to see the changes.
However, this is somewhat different in the case you have a const reference. Since you'll only ever read from a const reference, it makes total sense to be able to use temporaries there. This is why the compiler will "hack" around it for you, and give a more permanent address to temporaries that you want to "turn" into const references.
So, the rule is that you cannot get a non-const reference to a temporary value. (This slightly changed with C++11, where we have a new type of references that serve this exact purpose, but methods are expected to deal with those in a special way.)
Thank you all for your answers :-)
Here I gather your good ideas ;-)
Answer
Return by value is not const. For example, we can call non-const member functions of return by value:
class C {
public:
int x;
void set (int n) { x = n; } // non-const function
};
C bar() { return C(); }
int main ()
{
bar.set(5); // OK
}
But C++ does not allow non-const references to temporary objects.
However C++11 allow non-const rvalue-references to temporary objects. ;-)
Explanation
class C {};
void foo (C& c) {}
C bar() { return C(); }
//bar() returns a temporary object
//temporary objects cannot be non-const referenced
int main()
{
//foo() wants a mutable reference (i.e. non-const)
foo( bar() ); // => compilation error
}
Three fixes
Change foo declaration
void foo (const C& c) {}
Use another object
int main()
{
C c;
foo( c = bar() );
}
Use C++11 rvalue-reference
void foo(C && c) {}
Moreover
To confirm temporary objects are const, this above source code fails for the same reason:
class C {};
void foo(C& c) {}
int main()
{
foo( C() );
}

C++ Return by reference

Say, i have a function which returns a reference and i want to make sure that the caller only gets it as a reference and should not receive it as a copy.
Is this possible in C++?
In order to be more clear. I have a class like this.
class A
{
private:
std::vector<int> m_value;
A(A& a){ m_value = a.m_value; }
public:
A() {}
std::vector<int>& get_value() { return m_value; }
};
int main()
{
A a;
std::vector<int> x = a.get_value();
x.push_back(-1);
std::vector<int>& y = a.get_value();
std::cout << y.size();
return 0;
}
Thanks,
Gokul.
You can do what you want for your own classes by making the class non copyable.
You can make an class non copyable by putting the copy constructor and operator= as private or protected members.
class C
{
private:
C(const C& other);
const C& operator=(const C&);
};
There is a good example of making a NonCopyable class here that you can derive from for your own types.
If you are using boost you can also use boost::noncopyable.
Alt solution:
Another solution is to have a void return type and make the caller pass their variable by reference. That way no copy will be made as you're getting a reference to the caller's object.
If your function returns a reference to an object that shouldn't have been copied, then your function already has done what it could do to prevent copying. If someone else calls your function and copies the return value, then either
it's an error the caller made, because the object should never be copied (in which case the return type probably shouldn't have been copyable in the first place), or
it's irrelevant for the caller because the function is only called once in a week (in which case you must not try to cripple your callers' code), or
it's a pretty dumb oversight on the side of the caller (in which case the error will be found by profiling).
For #1, either you return have your own type or you can wrap whatever your return in your own type. Note that the only difference between #2 and #3 is the relevance - and if it's relevant, profiling will find it.
IMO you should not cripple your code by returning a pointer when what you need is a reference. Experienced programmers, seeing the pointer, will immediately ask whether they need to check for a NULL return value, whether the object is allocated dynamically and, if so, who is responsible for cleaning it up.
You should also not blindly forbid copying of whatever you return, if you cannot eliminate the possibility that copying is needed.
In the end it's the old motto, which C++ inherited from C: Trust your users to know what they are doing.
It "depends". Yes, you can hide the copy-constructor (and assignment operator), and your object becomes noncopyable:
struct foo
{
private:
foo(const foo&); // dont define
foo& operator=(const foo&); // dont define
}:
But if you're wondering about one specific function (i.e., normally copyable, but not for this function), no. In fact, what can you do about the caller anyway?
const foo& f = get_foo(); // okay, by reference, but...
foo f2 = foo(foo(foo(foo(foo(foo(f)))))); // :[
If your caller wants to do something, there isn't much you can do to stop it.
In C++11, you can prevent the copy constructor from being called by deleting it:
class A{
public:
A(const A&) = delete;
}
Are you trying to prevent a common typo that causes large objects to accidentally be copied? If so, you could return by pointer instead. Leaving off an & is pretty easy, but it takes a little bit of effort to copy an object from a pointer. OTOH, the resulting code will be uglier, so it's up to you whether it's worth it.