Consider this code:
struct A
{
A() = default;
A(const A&) = default;
};
If I'm correct, the compiler will automatically generated a default deconstructor. However, it will not do so for the movement constructor.
However, the above code allows me to do this:
int main()
{
A a1;
A a2 = std::move(a1);
return 0;
}
My question is: why?
If I modify the definition of struct A as follows:
struct A
{
A() = default;
A(const A&) = default;
A(A&&) = delete;
};
then A a2 = std::move(a1) does not compile, which makes sense.
So, what's the difference between deleting a constructor and not generating it? Or does the compiler generate the move constructor anyway?
Related
I'm working with an expression template class which should not be instantiated to avoid dangling references. But I'm temped to declare a variable with auto and 'auto' create a named instance of a temporary class.
How can I disable auto declaration of temporary class in the following code?
class Base
{
};
class Temp : public Base
{
public:
Temp() {}
Temp(int, int) {}
Temp(const Temp&) = default;
Temp(Temp&&) = default;
};
Temp testFunc(int a, int b) {
return Temp{a,b};
}
int main() {
Base a = testFunc(1,2); // this should work
auto b = testFunc(1,2); // this should fail to compile
return 0;
}
You seem to want to prevent users from using auto on a particular type. That's not possible in any version of C++. If it is legal C++ for a user to write T t = <expr>;, where T is the type of <expr>, then it will be legal for a user to write auto t = <expr>; (ignoring class data members). Just as you cannot forbid someone from passing <expr> to a template function using template argument deduction.
Anything you do to prevent auto usage will also inhibit some other usage of the type.
One option would be to make Temp's constructors private, move testFunc inside the Temp class and make it static. This way you can still instantiate Base, but auto would fail because you would be calling a private constructor:
class Base
{
};
class Temp : public Base
{
Temp() {}
Temp(int, int) {}
Temp(const Temp&) = default;
Temp(Temp&&) = default;
public:
static Temp testFunc(int a, int b)
{
return Temp{a,b};
}
};
int main() {
Base a = Temp::testFunc(1,2); // this should work
auto b = Temp::testFunc(1,2); // this should fail to compile
return 0;
}
Demo
I have the following example:
#include <cstdint>
class A
{
public:
A(const A&) = delete;
A& operator = (const A&) = delete;
A(A&&) = default; // why do I need the move constructor
A&& operator = (A&&) = delete;
explicit A(uint8_t Val)
{
_val = Val;
}
virtual ~A() = default;
private:
uint8_t _val = 0U;
};
class B
{
public:
B(const B&) = delete;
B& operator = (const B&) = delete;
B(B&&) = delete;
B&& operator = (B&&) = delete;
B() = default;
virtual ~B() = default;
private:
A _a = A(4U); // call the overloaded constructor of class A
};
int main()
{
B b;
return 0;
}
why do I need the move-constructor "A(A&&) = default;" in A? I could not the code line where the mentioned move-constructor is called.
Many thanks in advance.
A _a = A(4U) in this case depends on the move constructor.
This type of initialization is called copy initialization, and according to the standard it may invoke the move constructor, see 9.3/14.
I am still little bit confused by returning a const reference. Probably, this has been already discussed, however let's have following code as I did not find the same:
#include <vector>
#include <iostream>
struct A
{
int dataSize;
std::vector<char> data;
};
class T
{
public:
T();
~T();
const A& GetData();
private:
A dataA;
};
T::T() : dataA{1}
{
}
T::~T()
{
}
const A& T::GetData()
{
return dataA;
}
void main()
{
T t;
A dataReceivedCpy = {};
dataReceivedCpy = t.GetData();
const A& dataReceivedRef = t.GetData();
std::cout << dataReceivedRef.dataSize << std::endl;
}
What exactly happens, when I call
dataReceivedCpy = t.GetData();
Is this correct? From my point of view, it is and a copy of the requested struct is made, am I right?
On the other hand,
const A& dataReceivedRef = t.GetData();
returns a reference to object member, it is correct, unless the T object is not destructed. Am I right?
Yes, your understanding sounds correct.
dataReceivedCpy = t.GetData();
calls a copy assignment operator to put a copy of t.dataA in dataReceivedCpy.
const A& dataReceivedRef = t.GetData();
does no copying and makes dataReceivedRef a reference to t.dataA. It is valid for the lifetime of t.
If I have two classes D1 and D2 that both derive from class Base, and I want to construct a particular one based on say, a boolean variable, there are various well known techniques, eg use a factory, or use smart pointers.
For example,
std::unique_ptr<Base> b;
if (flag)
{
b.reset(new D1());
}
else
{
b.reset(new D2());
}
But this uses the heap for allocation, which is normally fine but I can think of times where it would be good to avoid the performance hit of a memory allocation.
I tried:
Base b = flag ? D1() : D2(); // doesn’t compile
Base& b = flag ? D1() : D2(); // doesn’t compile
Base&& b = flag ? D1() : D2(); // doesn’t compile
Base&& b = flag ? std::move(D1()) : std::move(D2()); // doesn’t compile
My intention is that D1 or D2 whichever is chosen is constructed on the stack, and its lifetime ends when b goes out of scope. Intuitively, I feel there should be a way to do it.
I played with lambda functions and found that this works:
Base&& b = [j]()->Base&&{
switch (j)
{
case 0:
return std::move(D1());
default:
return std::move(D2());
}
}();
Why it doesn’t suffer from the same issues as the others that do not compile I do not know.
Further, it would only be suitable for classes that are inexpensive to copy, because despite my explicit request to use move, it does I think still call a copy constructor. But if I take away the std::move, I get a warning!
I feel this is closer to what i think should be possible but it still has some issues:
the lambda syntax is not friendly to old-timers who havent yet
embraced the new features of the language ( myself included)
the copy constructor call as mentioned
Is there a better way of doing this?
If you know all the types, you can use a Boost.Variant, as in:
class Manager
{
using variant_type = boost::variant<Derived1, Derived2>;
struct NameVisitor : boost::static_visitor<const char*>
{
template<typename T>
result_type operator()(T& t) const { return t.name(); }
};
public:
template<typename T>
explicit Manager(T t) : v_(std::move(t)) {}
template<typename T>
Manager& operator=(T t)
{ v_ = std::move(t); return *this; }
const char* name()
{ return boost::apply_visitor(NameVisitor(), v_); }
private:
variant_type v_;
};
Note: by using variant, you no longer need a base class or virtual functions.
The way you are trying to do it, you are going to get a dangling reference. Having the std::move is just hiding that.
Generally I just structure the code so that the logic is in a separate function. That is, instead of
void f(bool flag)
{
Base &b = // some magic to choose which derived class to instantiate
// do something with b
}
I do
void doSomethingWith(Base &b)
{
// do something with b
}
void f(bool flag)
{
if (flag) {
D1 d1;
doSomethingWith(d1);
}
else {
D2 d2;
doSomethingWith(d2);
}
}
However, if that doesn't work for you, you can use a union inside a class to help manage it:
#include <iostream>
using std::cerr;
struct Base {
virtual ~Base() { }
virtual const char* name() = 0;
};
struct Derived1 : Base {
Derived1() { cerr << "Constructing Derived1\n"; }
~Derived1() { cerr << "Destructing Derived1\n"; }
virtual const char* name() { return "Derived1"; }
};
struct Derived2 : Base {
Derived2() { cerr << "Constructing Derived2\n"; }
~Derived2() { cerr << "Destructing Derived2\n"; }
virtual const char* name() { return "Derived2"; }
};
template <typename B,typename D1,typename D2>
class Either {
union D {
D1 d1;
D2 d2;
D() { }
~D() { }
} d;
bool flag;
public:
Either(bool flag)
: flag(flag)
{
if (flag) {
new (&d.d1) D1;
}
else {
new (&d.d2) D2;
}
}
~Either()
{
if (flag) {
d.d1.~D1();
}
else {
d.d2.~D2();
}
}
B& value()
{
if (flag) {
return d.d1;
}
else {
return d.d2;
}
}
};
static void test(bool flag)
{
Either<Base,Derived1,Derived2> either(flag);
Base &b = either.value();
cerr << "name=" << b.name() << "\n";
}
int main()
{
test(true);
test(false);
}
gives this output:
Constructing Derived1
name=Derived1
Destructing Derived1
Constructing Derived2
name=Derived2
Destructing Derived2
You can ensure you have enough space for allocating either on the stack with std::aligned_storage. Something like:
// use macros for MAX since std::max is not const-expr
std::aligned_storage<MAX(sizeof(D1), sizeof(D2)), MAX(alignof(D1), alignof(D2))> storage;
Base* b = nullptr;
if (flag)
b = new (&storage) D1();
else
b = new (&storage) D2();
You can make a wrapper type for aligned_storage that just takes two types and does the maximum of size/alignment of the two without needing to repeat yourself in the code using it. You can emulate aligned_storage for non-over-aligned types fairly trivially too if you need C++98 support. The custom type without over-aligned support would be something like:
template <typename T1, typename T2>
class storage
{
union
{
double d; // to force strictest alignment (on most platforms)
char b[sizeof(T1) > sizeof(T2) ? sizeof(T1) : sizeof(T2)];
} u;
};
And that can be given protections against copies/moves if you so wish. It can even be turned into a simplified Boost.Variant with relatively little work.
Note that with this approach (or some of the others), destructors will not be called automatically on your class and you must call them yourself. If you want RAII patterns to apply here, you can extend the example class above to store a deleter function that is bound during construction into the space.
template <typename T1, typename T2>
class storage
{
using deleter_t = void(*)(void*);
std::aligned_storage<
sizeof(T1) > sizeof(T2) ? sizeof(T1) : sizeof(T2),
alignof(T1) > alignof(T2) ? alignof(T1) : alignof(T2)
> space;
deleter_t deleter = nullptr;
public:
storage(const storage&) = delete;
storage& operator=(const storage&) = delete;
template <typename T, typename ...P>
T* emplace(P&&... p)
{
destroy();
deleter = [](void* obj){ static_cast<T*>(obj)->~T(); }
return new (&space) T(std::forward<P>(p)...);
}
void destroy()
{
if (deleter != nullptr)
{
deleter(&space);
deleter = nullptr;
}
}
};
// usage:
storage<D1, D2> s;
B* b = flag ? s.emplace<D1>() : s.emplace<D2>();
And of course that can all be done in C++98, just with a lot more work (especially in terms of emulating the emplace function).
How about
B&&b = flag ? static_cast<B&&>(D1()) : static_cast<B&&>(D2());
If you just need them to be freed when the reference goes out of scope, you could implement another simple class (maybe named DestructorDecorator) that points to the object (D1 or D2). And then you just have to implement ~DestructorDecorator to call the destructor of D1 or D2.
You haven't mentioned it, your flag is known at compile time?
As far as a compile-time flag is concerned, you can use template magic to deal with the conditional construction of the class:
First, declaring a template create_if which takes two types and a boolean:
template <typename T, typename F, bool B> struct create_if {};
Second, specializing create_if for true and false values:
template <typename T, typename F> struct create_if<T, F, true> { using type = T; };
template <typename T, typename F> struct create_if<T, F, false> { using type = F; };
Then, you can do this:
create_if<D1, D2, true>::type da; // Create D1 instance
create_if<D1, D2, false>::type db; // Create D2 instance
You can change the boolean literals with your compile-time flag or with a constexpr function:
constexpr bool foo(const int i) { return i & 1; }
create_if<D1, D2, foo(100)>::type dc; // Create D2 instance
create_if<D1, D2, foo(543)>::type dd; // Create D1 instance
This is valid only if the flag is known at compile time, I hope it helps.
Live example.
I'm trying to use move semantics (just as an experiment).
Here is my code:
class MyClass {
public:
MyClass(size_t c): count(c) {
data = new int[count];
}
MyClass( MyClass&& src) : count(src.count) {
data = src.data;
src.count = 0;
src.data = nullptr;
}
void operator=( MyClass&& src) {
data = src.data;
count = src.count;
src.count = 0;
src.data = nullptr;
}
~MyClass() {
if (data != nullptr)
delete[] data;
}
int* get_data() const {
return data;
}
size_t get_count() const {
return count;
}
private:
MyClass(const MyClass& src) : count(src.count) {
data = new int[src.count];
memcpy(data, src.data, sizeof(int)*src.count);
}
void operator=(const MyClass& src) {
count = src.count;
data = new int[src.count];
memcpy(data, src.data, sizeof(int)*src.count);
}
int* data;
size_t count;
};
int main()
{
MyClass mc(150);
for (size_t i = 0; i < mc.get_count(); ++i)
mc.get_data()[i] = i;
MyClass &&mc2 = std::move(mc);
return 0;
}
But std::move does not move mc to mc2, it just copies (copyies pointer as it is). If I remove copy constructor compiler generates it for MyClass.
How can I force move semantics to be used? How can I make it to be used in such constructions:
MyClass mc2(mc); //Move, not copy
-or-
MyClass mc2 = mc; //Move, not copy
I tried to use a '&&' operator to explicitely mark rvalue, but, of cause, it didn't work.
You're declaring m2 as a reference, not as a value. So it still refers to what it was initialised with, namely m1. You wanted this:
MyClass mc2 = std::move(mc);
Live example
As for the second part - there is no way to force a construct like these:
MyClass mc2(mc); //Move, not copy
//-or-
MyClass mc2 = mc; //Move, not copy
to move. If you want to move from an lvalue (and mc is indeed an lvalue), you have to use std::move (or another cast to rvalue) explicitly.
There is one thing you could do, but it would be a dirty hack, make the code unintuitive and be a great source for bugs. You could add an overload of the copy constructor (and copy assignment operator) taking a non-const reference, which would do the move. Basically something like std::auto_ptr used to do before it was rightfully deprecated. But it would never pass code review with me, for example. If you want to move, just std::move.
A few side notes:
Calling delete or delete[] on a null pointer is guaranteed to be a no-op, so you can safely drop the if from your destructor.
It's generally preferable to use std::copy instead of memcpy in C++ code, you don't have to worry about getting the sizeof right
You can force move semantics, if you delete the copy constructor and the assignment operator
MyClass(const MyClass& src)= delete;
void operator=(const MyClass& src) = delete;
in this case the provided move constructor or move assignment operator will be picked.
Rewrite your class a bit with some comments. Look over it, you might notice a few things you missed. Like:
in MyClass(size_t c) not checking for c != 0.
in void operator=(const MyClass& src) not delete[] data; (if exists) before reallocating.
And some other tiny details.Hope your compiler can handle this.
class MyClass {
private:
// initialize memebers directly
int* data = nullptr;
size_t count = 0;
public:
// default empty contructor
MyClass() = default;
// destructor
~MyClass() {
*this = nullptr; // use operator = (nullptr_t)
}
// allow nullptr construct
MyClass(nullptr_t):MyClass() {}
// allow nullptr assignment (for clearing)
MyClass& operator = (nullptr_t) {
if(data) {
delete[] data;
data = nullptr;
}
count = 0;
return *this;
}
// chain to default constructor, redundant in this case
MyClass(size_t c):MyClass() {
// maybe size_t is 0?
if(count = c) {
data = new int[count];
}
}
// chain to default constructor, redundant in this case
MyClass(MyClass&& src):MyClass() {
*this = std::move(src); // forward to move assignment
}
MyClass& operator=(MyClass&& src) {
// don't swap with self
if(&src != this) {
// it's better to swap and let src destroy when it feels like it.
// I always write move contructor and assignment to swap data.
// it's gonna be destroyed anyway, or not...
std::swap(src.data, data);
std::swap(src.count, count);
}
return *this;
}
MyClass(const MyClass& src):MyClass() {
*this = src; // forward to copy assignment
}
MyClass& operator = (const MyClass& src) {
// don't copy to self
if(&src != this) {
// delete first
if(data) {
delete[] data;
data = nullptr;
}
// now reallocate
if(count = src.count) {
data = new int[count];
memcpy(data, src.data, sizeof(int)* count);
}
}
return *this;
}
// easy way to use the object in a if(object) to test if it has content
explicit operator bool() const {
return data && count;
}
// same as above but made for if(!object) to test if empty
bool operator !() const {
return !data || !count;
}
public:
int* get_data() const {
return data;
}
size_t get_count() const {
return count;
}
// add more custom methods
};
Now to move you do this:
MyClass object1; // default construct
MyClass object1(5); // construct with capacity
MyClass object2(object1); // copy constructor
MyClass object3(std::move(object1)); // move constructor
object2 = object1; // copy assignment
object3 = std::move(object1); // move constructor
std::swap(object2, object3); // swap the two
object2 = nullptr; // to empty it
if(object1); // bool cast