The following code will not compile on gcc 4.8.2.
The problem is that this code will attempt to copy construct an std::pair<int, A> which can't happen due to struct A missing copy and move constructors.
Is gcc failing here or am I missing something?
#include <map>
struct A
{
int bla;
A(int blub):bla(blub){}
A(A&&) = delete;
A(const A&) = delete;
A& operator=(A&&) = delete;
A& operator=(const A&) = delete;
};
int main()
{
std::map<int, A> map;
map.emplace(1, 2); // doesn't work
map.emplace(std::piecewise_construct,
std::forward_as_tuple(1),
std::forward_as_tuple(2)
); // works like a charm
return 0;
}
As far as I can tell, the issue isn't caused by map::emplace, but by pair's constructors:
#include <map>
struct A
{
A(int) {}
A(A&&) = delete;
A(A const&) = delete;
};
int main()
{
std::pair<int, A> x(1, 4); // error
}
This code example doesn't compile, neither with coliru's g++4.8.1 nor with clang++3.5, which are both using libstdc++, as far as I can tell.
The issue is rooted in the fact that although we can construct
A t(4);
that is, std::is_constructible<A, int>::value == true, we cannot implicitly convert an int to an A [conv]/3
An expression e can be implicitly converted to a type T if and only if the declaration T t=e; is well-formed,
for some invented temporary variable t.
Note the copy-initialization (the =). This creates a temporary A and initializes t from this temporary, [dcl.init]/17. This initialization from a temporary tries to call the deleted move ctor of A, which makes the conversion ill-formed.
As we cannot convert from an int to an A, the constructor of pair that one would expect to be called is rejected by SFINAE. This behaviour is surprising, N4387 - Improving pair and tuple analyses and tries to improve the situation, by making the constructor explicit instead of rejecting it. N4387 has been voted into C++1z at the Lenexa meeting.
The following describes the C++11 rules.
The constructor I had expected to be called is described in [pairs.pair]/7-9
template<class U, class V> constexpr pair(U&& x, V&& y);
7 Requires: is_constructible<first_type, U&&>::value is true and
is_constructible<second_type, V&&>::value is true.
8 Effects: The
constructor initializes first with std::forward<U>(x) and second with
std::forward<V>(y).
9 Remarks: If U is not implicitly convertible to
first_type or V is not implicitly convertible to second_type this
constructor shall not participate in overload resolution.
Note the difference between is_constructible in the Requires section, and "is not implicitly convertible" in the Remarks section. The requirements are fulfilled to call this constructor, but it may not participate in overload resolution (= has to be rejected via SFINAE).
Therefore, overload resolution needs to select a "worse match", namely one whose second parameter is a A const&. A temporary is created from the int argument and bound to this reference, and the reference is used to initialize the pair data member (.second). The initialization tries to call the deleted copy ctor of A, and the construction of the pair is ill-formed.
libstdc++ has (as an extension) some nonstandard ctors. In the latest doxygen (and in 4.8.2), the constructor of pair that I had expected to be called (being surprised by the rules required by the Standard) is:
template<class _U1, class _U2,
class = typename enable_if<__and_<is_convertible<_U1, _T1>,
is_convertible<_U2, _T2>
>::value
>::type>
constexpr pair(_U1&& __x, _U2&& __y)
: first(std::forward<_U1>(__x)), second(std::forward<_U2>(__y)) { }
and the one that is actually called is the non-standard:
// DR 811.
template<class _U1,
class = typename enable_if<is_convertible<_U1, _T1>::value>::type>
constexpr pair(_U1&& __x, const _T2& __y)
: first(std::forward<_U1>(__x)), second(__y) { }
The program is ill-formed according to the Standard, it is not merely rejected by this non-standard ctor.
As a final remark, here's the specification of is_constructible and is_convertible.
is_constructible [meta.rel]/4
Given the following function prototype:
template <class T>
typename add_rvalue_reference<T>::type create();
the predicate condition for a template specialization is_constructible<T, Args...> shall be satisfied if and only if the following variable definition would be well-formed for some invented variable t:
T t(create<Args>()...);
[Note: These tokens are never interpreted as a function declaration. β end note] Access checking is performed as if in a context unrelated to T and any of the Args. Only the validity of the immediate context of the variable initialization is considered.
is_convertible [meta.unary.prop]/6:
Given the following function prototype:
template <class T>
typename add_rvalue_reference<T>::type create();
the predicate condition for a template specialization is_convertible<From, To> shall be satisfied if and
only if the return expression in the following code would be well-formed, including any implicit conversions
to the return type of the function:
To test() {
return create<From>();
}
[Note: This requirement gives well defined results for reference types, void types, array types, and function types. β end note] Access checking is performed as if in a context unrelated to To and From. Only
the validity of the immediate context of the expression of the return-statement (including conversions to
the return type) is considered.
For your type A,
A t(create<int>());
is well-formed; however
A test() {
return create<int>();
}
creates a temporary of type A and tries to move that into the return-value (copy-initialization). That selects the deleted ctor A(A&&) and is therefore ill-formed.
Related
For the following code snippet (using C++14 standard), can we declare setDataVectoras noexcept?
class Data {
public:
using Type = ...; // A class with a default move assignment operator. Or even just uint32_t
void setDataVector(std::vector<Type> &&input) // Can be declared as noexcept?
{
data = std::move(input);
}
private:
std::vector<Type> data;
};
In cppreference, it is mentioned that until C++17, the move assignment operator for std::vector is not noexcept.
vector& operator=( vector&& other );
Can it really throw even for trivial datatypes like integer?
What is confusing me is that move operations should be exception-safe but std::vector and std::string for example don't have a noexcept move assignment operators.
So, What am I missing here?
Simplified code snippet is:
class A {
public:
~A();
static A create();
private:
A() = default;
A(A&&) = default;
NonCopyable n;
};
A A::create() {
A a;
return a;
}
int main(int argc, char* argv[]) {
auto a = A::create();
return 0;
}
Please also see my live example (which shows different compilers' behavior).
In the end, I'm wondering why does auto a = A::create(); compile without errors using newer compilers [gcc >= 7.1] (which part of the C++17 standard is relevant here?), given that:
We have a non-copyable member NonCopyable n;, so default copy constructor would be ill-formed.
It's an NRVO here since A a; return a; so copy elision is not guaranteed by the standard.
Move constructor A(A&&) is marked private.
Optimizations were off -O0 for testing.
My suspicion is that move constructor is being "validated" by the compiler at return a;; since this is a member function of A it passes the validation. Even if the suspicion is correct, I'm not sure if this is standard-compliant.
I believe this is a consequence of P0135: Wording for guaranteed copy elision through simplified value categories, specifically the change to [dcl.init]:
If the initializer expression is a prvalue and the cv-unqualified version of the source type is the same class as the class of the destination, the initializer expression is used to initialize the destination object.
[Example: T x = T(T(T())); calls the T default constructor to initialize x. ββend example]
As a result, this behavior is not dependent on copy elision of return values or the availability of move constructors.
Given the following struct:
struct ABC
{
ABC(){cout << "ABC" << endl;}
~ABC() noexcept {cout << "~ABC" << endl;}
ABC(ABC const&) {cout << "copy" << endl;}
ABC(ABC&&) noexcept {cout << "move" << endl;}
ABC& operator=(ABC const&){cout << "copy=" << endl;}
ABC& operator=(ABC&&) noexcept {cout << "move=" << endl;}
};
The output of:
std::pair<std::string, ABC> myPair{{}, {}};
is:
ABC
copy
~ABC
~ABC
While the output of:
std::pair<std::string, ABC> myPair{{}, ABC{}};
is:
ABC
move
~ABC
~ABC
In attempting to understand the difference between the two I think I have identified that the first case is using copy-list-initialization, while the second one uses direct-list-initialization of an unnamed temporary (numbers 7 and 2, respectively, in here: http://en.cppreference.com/w/cpp/language/list_initialization).
Searching for similar questions I've found this: Why does the standard differentiate between direct-list-initialization and copy-list-initialization? and this: Does copy list initialization invoke copy ctor conceptually?.
The answers in those questions discuss the fact that for copy-list-initialization, the use of an explicit constructor would render the code ill-formed. In fact, if I make ABC's default constructor explicit, my first example won't compile but that is (perhaps) a different matter.
So, the question is: Why is the temporary copied in the first case but moved in the second? What prevents it from being moved in the case of copy-list-initialization?
As a note, the following code:
std::pair<std::string, ABC> myPair = std::make_pair<string, ABC>({}, {});
Also results in a call to ABC's move constructor (and no copy constructor call), but different mechanisms may be involved.
You can try the code out (using gcc-4.9.2 in C++14 mode) at: https://ideone.com/Kc8xIn
In general, braced-init-lists like {} are not expressions and do not have a type. If you have a function template
template<typename T> void f(T);
and call f( {} ), no type will be deduced for T, and type deduction will fail.
On the other hand, ABC{} is a prvalue expression of type ABC (an "explicit type conversion in functional notation"). For a call like f( ABC{} ), the function template can deduce the type ABC from this expression.
In C++14, as well as in C++11, std::pair has the following constructors [pairs.pair]; T1 and T2 are the names of the template parameter of the std::pair class template:
pair(const pair&) = default;
pair(pair&&) = default;
constexpr pair();
constexpr pair(const T1& x, const T2& y);
template<class U, class V> constexpr pair(U&& x, V&& y);
template<class U, class V> constexpr pair(const pair<U, V>& p);
template<class U, class V> constexpr pair(pair<U, V>&& p);
template <class... Args1, class... Args2>
pair(piecewise_construct_t, tuple<Args1...>, tuple<Args2...>);
Note that there is a constructor
constexpr pair(const T1& x, const T2& y); // (C)
But no
constexpr pair(T1&& x, T2&& y);
instead, there is a perfectly forwarding
template<class U, class V> constexpr pair(U&& x, V&& y); // (P)
If you try to initialize a std::pair with two initializers where at least one of them is a braced-init-list, the constructor (P) is not viable since it cannot deduce its template arguments.
(C) is not a constructor template. Its parameter types T1 const& and T2 const& are fixed by the class template parameters. A reference to a constant type can be initialized from an empty braced-init-list. This creates a temporary object that is bound to the reference. As the type referred to is const, the (C) constructor will copy its arguments into the class' data members.
When you initialize a pair via std::pair<T,U>{ T{}, U{} }, the T{} and U{} are prvalue-expressions. The constructor template (P) can deduce their types and is viable. The instantiation produced after type deduction is a better match than the (C) constructor, because (P) will produce rvalue-reference parameters and bind the prvalue arguments to them. (C) on the other hand binds the prvalue arguments to lvalue-references.
Why then does the live example move the second argument when called via std::pair<T,U>{ {}, U{} }?
libstdc++ defines additional constructors. Below is an extract of its std::pair implementation from 78536ab78e, omitting function definitions, some comments, and SFINAE. _T1 and _T2 are the names of the template parameters of the std::pair class template.
_GLIBCXX_CONSTEXPR pair();
_GLIBCXX_CONSTEXPR pair(const _T1& __a, const _T2& __b); // (C)
template<class _U1, class _U2>
constexpr pair(const pair<_U1, _U2>& __p);
constexpr pair(const pair&) = default;
constexpr pair(pair&&) = default;
// DR 811.
template<class _U1>
constexpr pair(_U1&& __x, const _T2& __y); // (X)
template<class _U2>
constexpr pair(const _T1& __x, _U2&& __y); // (E) <=====================
template<class _U1, class _U2>
constexpr pair(_U1&& __x, _U2&& __y); // (P)
template<class _U1, class _U2>
constexpr pair(pair<_U1, _U2>&& __p);
template<typename... _Args1, typename... _Args2>
pair(piecewise_construct_t, tuple<_Args1...>, tuple<_Args2...>);
Note the (E) constructor template: It will copy the first argument and perfectly forward the second. For an initialization like std::pair<T,U>{ {}, U{} }, it is viable because it only needs to deduce a type from the second argument. It is also a better match than (C) for the second argument, and hence a better match overall.
The "DR 811" comment is in the libstdc++ sources. It refers to LWG DR 811 which adds some SFINAE, but no new constructors.
The constructors (E) and (X) are a libstdc++ extension. I'm not sure if it's compliant, though.
libc++ on the other hand does not have this additional constructors. For the example std::pair<T,U>{ {}, U{} }, it will copy the second argument.
Live demo with both library implementations
I ran into a problem. I implemented a constructor for a class, but why are implicitly generated the other constructors, like the copy one?
I thought, that if I define a constructor explicitly, then the compiler doesn't generates implicitly other ones. I'm really hoping, that this a VC++ specific thing, and that this code doesn't conforms to ISO:IEC C++11:
class Foo
{
int bar;
public:
Foo(int&& arg) : bar(arg) { cout << "RConstruction" << endl; }
};
int main(int, const char*[])
{
Foo f = Foo(42);
/* Create unused temporary on the stack */
Foo::Foo(f); /* calling Foo::Foo(const Foo&): this shouldn't work... */
return (0);
}
Please keep in mind, that this is a sample code, created exactly for this situation, for demonstration purposes, I expect answers only that strictly relate to this question.
That's not a move constructor, so it doesn't suppress any implicit ones.
Just like Foo(const int&) isn't a copy constructor, Foo(int&&) isn't a move constructor, it's just a constructor taking an rvalue reference.
A move constructor looks like one of:
Foo(Foo&&)
Foo(const Foo&&)
Foo(volatile Foo&&)
Foo(const volatile Foo&&)
I thought, that if I define a constructor explicitly, then the compiler doesn't generates implicitly other ones.
If you define any constructor the compiler doesn't generate the default constructor, but it still generates the others. Define the as deleted if you don't want them:
Foo(const Foo&) = delete;
Foo& operator=(const Foo&) = delete;
You did not declare a move constructor, but a regular constructor : no implicit constructor will be deleted.
A move constructor would be of the form Foo(Foo&& arg) (with any cv-qualifier on the argument)
Also note that this statement is not valid C++ :
Foo::Foo(f);
Maybe you meant :
Foo g = Foo(f);
I have the boost::variant over set of non-default constructible (and maybe even non-moveable/non-copyable and non-copy/move constructible) classes with essentialy different non-default constructor prototypes, as shown below:
#include <boost/variant.hpp>
#include <string>
#include <list>
struct A { A(int) { ; } };
struct B { B(std::string) { ; } };
struct C { C(int, std::string) { ; } };
using V = boost::variant< A const, B const, C const >;
using L = std::list< V >;
int main()
{
L l;
l.push_back(A(1)); // an extra copy/move operation
l.push_back(B("2")); // an extra copy/move operation
l.push_back(C(3, "3")); // an extra copy/move operation
l.emplace_back(4);
l.emplace_back(std::string("5"));
// l.emplace_back(3, std::string("3")); // error here
return 0;
}
I expect, that std::list::emplace_back allows me to construct-and-insert (in single operation) new objects (of all the A, B, C types) into list, even if they have T & operator = (T const &) = delete;/T & operator = (T &&) = delete; and T(T const &) = delete;/T(T &&) = delete;. But what should I do, if constructor is a non-conversion one? I.e. have more, than one parameter. Or what I should to do if two different variant's underlying types have ambiguous constructor prototypes? In my opinion, this is the defect of implementation of the boost::variant library in the light of the new features of C++11 standard, if any at all can be applyed to solve the problem.
I specifically asked about std::list and boost::variant in superposition, because they are both internally implement the pimpl idiom in some form, as far as I know (say, boost::variant currently designed by means of temporary heap backup approach).
emplace can only call the constructors of the type in question. And boost::variant's constructors only take single objects which are unambiguously convertible to one of the variant's types.
variant doesn't forward parameters arbitrarily to one of its bounded types. It just takes a value. A single value that it will try to convert to one of the bounded types.
So you're going to have to construct an object and then copy that into the variant.
Assuming you can modify your "C" class, you could give it an additional constructor that takes a single tuple argument.