Is there a way to avoid copying when initializing a vector?
The code below will produce the following output.
#include <iostream>
#include <vector>
using namespace std;
struct Ticker {
std::string m_ticker;
Ticker() {
std::cout << "Default constructor" << std::endl;
}
Ticker(const std::string& ticker)
: m_ticker(ticker)
{
std::cout << "Parametrized constructor" << std::endl;
}
Ticker(Ticker&& other)
{
std::cout << "Move constructor" << std::endl;
m_ticker = other.m_ticker;
other.m_ticker = "";
}
Ticker(const Ticker& x)
{
std::cout << "Copy constructor" << std::endl;
m_ticker = x.m_ticker;
}
~Ticker()
{
std::cout << "Destructor" << std::endl;
}
friend std::ostream& operator << (std::ostream& os, const Ticker& dr);
};
std::ostream& operator << (std::ostream& os, const Ticker& dr)
{
os << "|" << dr.m_ticker << "|";
return os;
}
int main() {
std::vector<Ticker> table = std::move(std::vector<Ticker>{std::move(Ticker("MSFT")), std::move(Ticker("TSL"))});
for (const auto& row: table)
{
std::cout << row << std::endl;
}
return 0;
}
This produces the following output:
Parametrized constructor
Move constructor
Parametrized constructor
Move constructor
Copy constructor
Copy constructor
Destructor
Destructor
Destructor
Destructor
|MSFT|
|TSL|
Destructor
Destructor
Is there a way to avoid the copy constructor and initialize in-place or just move without copying?
If you use
std::vector<Ticker> table = std::vector<Ticker>{Ticker("MSFT"), Ticker("TSL")};
You will get
Parametrized constructor
Parametrized constructor
Copy constructor
Copy constructor
Destructor
Destructor
|MSFT|
|TSL|
Destructor
Destructor
Which has 4 constructor calls instead of the 6 you currently have. 2 of those calls are for Ticker("MSFT") and Ticker("TSL") and then the additional two copies are because initializer lists store the elements in them as const, so they have to be copied into the vector as you can't move from a const object.
To get the bare minimum of 2 constructor calls you'll need to use the emplace_back member function like
std::vector<Ticker> table; // create empty vector
table.reserve(2); // allocate space for 2 Tickers but create nothing
table.emplace_back("MSFT"); // directly construct from "MSFT" in the reserved space
table.emplace_back("TSL"); // directly construct from "TSL" in the reserved space
which has the output of
Parametrized constructor
Parametrized constructor
|MSFT|
|TSL|
Destructor
Destructor
If you want a syntax like std::vector<Ticker> table = std::vector<Ticker>{Ticker("MSFT"), Ticker("TSL")};, but without the extra overhead, you could wrap the emplace_back solution in a factory function like
template <typename T, typename... Args>
auto make_vector(Args&&... args)
{
std::vector<T> data;
data.reserve(sizeof...(Args));
(data.emplace_back(std::forward<Args>(args)), ...);
return data;
}
and then you would use it like
auto table = make_vector<Ticker>("MSFT", "TSL");
Related
Consider the following C++-Code
#include <iostream>
using namespace std;
struct WrapMe
{
WrapMe() { cout << "WrapMe Default Ctor of: " << this << endl; }
WrapMe(const WrapMe& other) { cout << "WrapMe Copy Ctor of " << this << " from " << &other << endl; }
WrapMe(WrapMe&& other) noexcept { cout << "WrapMe Move Ctor of " << this << " from " << &other << endl; }
~WrapMe() { cout << "Wrap Me Dtor of" << this << endl; }
};
struct Wrapper1
{
WrapMe& data()& { return member; }
WrapMe data()&& { return std::move(member); }
WrapMe member;
};
struct Wrapper2
{
WrapMe& data()& { return member; }
WrapMe&& data()&& { return std::move(member); }
WrapMe member;
};
int main()
{
auto wrapMe1 = Wrapper1().data();
auto wrapMe2 = Wrapper2().data();
}
with the output
WrapMe Default Ctor of: 00000092E7F2F8C4
WrapMe Move Ctor of 00000092E7F2F7C4 from 00000092E7F2F8C4
Wrap Me Dtor of00000092E7F2F8C4
WrapMe Default Ctor of: 00000092E7F2F8E4
WrapMe Move Ctor of 00000092E7F2F7E4 from 00000092E7F2F8E4
Wrap Me Dtor of00000092E7F2F8E4
[...]
Which is the correct way to move from the WrapMe member: Like Wrapper1 (return by value) or like Wrapper2 (return by rvalue-reference) does? Or are both ways equivalent here, as the ouput suggests? If not, why?
WrapMe&& data()&& { return std::move(member); }
This doesn't actually move anything. It just returns a rvalue reference to the member. For example I could do
auto&& wrapMe2 = Wrapper2().data();
and now wrapMe2 will be a dangling reference or
auto w = wrapper2();
auto&& wrapMe2 = std::move(x).data();
Now I have a reference to the member of w without w having changed at all.
Only because the move constructor of WrapMe is called to initialize the wrapMe2 object in the original line, a move operations actually takes place.
The version
WrapMe data()&& { return std::move(member); }
calls the move constructor already to construct the return value. The returned value will never refer to the original object or its member.
The line
auto wrapMe1 = Wrapper1().data();
then calls the move constructor again to initialize wrapMe1 from the return value of .data(). The compiler is however allowed to elide this second move constructor call and instead construct wrapMe1 directly from the expression in the return statement. This is why you see the same result.
With C++17 or later this elision would even be mandatory.
I don't know your use case, so I cannot be sure what the correct approach is, but just guessing on what you want to do:
For consistency between the two overloads, I would use the reference-returning version. Having data provide modifiable access to the member for lvalues, but not for rvalues, would be confusing.
However, you don't really need a member function to do this. Through data the caller has full control over the member anyway, so you could just make the member public from the start and then it could be used in the same way directly.
Currently, I'm storing a collection of std::unique_ptrs to heap allocated objects of a polymorphic type:
struct Foo {
virtual ~Foo() = default;
};
Collection<std::unique_ptr<Foo>> foos;
The basic interface I need is putting / taking owners of Foo to / from foos. The objects stored in foos are never supposed to be nullptr so I'd like to replace runtime assert(owner_taken) with compile-time checks. Moreover, I would like to be capable of using non-null owners in the context where a nullable one may be expected.
Probably, I need to store something like unique_ref but how would I extract one from foos? I don't want a copy, I want the stored object itself, so owner->clone() isn't a solution. Neither I can std::move(owner) because the state of a "unique reference" would be invalid afterwards.
Is there a clean design decision?
Is there a never-null unique owner of heap allocated objects?
There is no such type in the standard library.
It is possible to implement such type. Simply define a type with unique_ptr member and mark the default constructor deleted. You can mark constructor from std::nullptr_t deleted also so that construction from nullptr will be prevented at compile time.
Whether you construct from an external pointer, or allocate in the constructor, there is no alternative for checking for null at runtime.
Reading your question, I interpret the following requirements:
You don't want to copy or move the object itself (Foo)
You don't want a wrapper which has some sort of empty state which excludes move semantics
The object itself (Foo) should be stored on the heap such that its lifetime is independent of the control flow
The object itself (Foo) should be deleted once it is not used any more
As construction / destruction, copy and move are the only ways to get objects into / out of a container, the only thing left is a wrapper object which is copied when moved into / out of the container.
You can create such an object yourself as follows:
// LICENSE: MIT
#include <memory>
#include <utility>
template<typename T>
class shared_ref {
public:
template<typename ...Args>
shared_ref(Args&&... args)
: data{new T(std::forward<Args>(args)...)}
{}
shared_ref(shared_ref&&) = delete;
shared_ref& operator=(shared_ref&&) = delete;
T& get() noexcept {
return *data;
}
const T& get() const noexcept {
return *data;
}
operator T&() noexcept {
return get();
}
operator const T&() const noexcept {
return get();
}
void swap(shared_ref& other) noexcept {
return data.swap(other);
}
private:
std::shared_ptr<T> data;
};
template<class T>
void swap(shared_ref<T>& lhs, shared_ref<T>& rhs) noexcept {
return lhs.swap(rhs);
}
I leave it as an exercise to the reader to add support for Allocator, Deleter, operator[], implicit conversion contructor to base classes.
This can then be used as follows:
#include <iostream>
int main() {
shared_ref<int> r; // default initialized
std::cout << "r=" << r << std::endl;
r = 5; // type conversion operator to reference
std::cout << "r=" << r << std::endl;
shared_ref<int> s{7}; // initialized with forwarded arguments
std::cout << "s=" << s << std::endl;
std::swap(r, s);
std::cout << "r=" << r << ", " << "s=" << s << std::endl;
s = r; // copy assignment operator
std::cout << "s=" << s << std::endl;
const shared_ref<int> t{s}; // copy constructor
std::cout << "t=" << t << std::endl;
//t = 8; // const ref from a const object is immutable
return 0;
}
I'm currently writing a logging class (just for practice) and ran into an issue. I have two classes: The class Buffer acts as a temporary buffer and flushes itself in it's destructor. And the class Proxy that returns a Buffer instance, so I don't have to write Buffer() all the time.
Anyways, here is the code:
#include <iomanip>
#include <iostream>
#include <sstream>
#include <string>
class Buffer
{
private:
std::stringstream buf;
public:
Buffer(){};
template <typename T>
Buffer(const T& v)
{
buf << v;
std::cout << "Constructor called\n";
};
~Buffer()
{
std::cout << "HEADER: " << buf.str() << "\n";
}
Buffer(const Buffer& b)
{
std::cout << "Copy-constructor called\n";
// How to get rid of this?
};
Buffer(Buffer&&) = default;
Buffer& operator=(const Buffer&) & = delete;
Buffer& operator=(Buffer&&) & = delete;
template <typename T>
Buffer& operator<<(const T& v)
{
buf << v;
return *this;
}
};
class Proxy
{
public:
Proxy(){};
~Proxy(){};
Proxy(const Proxy&) = delete;
Proxy(Proxy&&) = delete;
Proxy& operator=(const Proxy&) & = delete;
Proxy& operator=(Proxy&&) & = delete;
template <typename T>
Buffer operator<<(const T& v) const
{
if(v < 0)
return Buffer();
else
return Buffer(v);
}
};
int main () {
Buffer(Buffer() << "Test") << "what";
Buffer() << "This " << "works " << "fine";
const Proxy pr;
pr << "This " << "doesn't " << "use the copy-constructor";
pr << "This is a " << std::setw(10) << " test";
return 0;
}
Here is the output:
Copy-constructor called
HEADER: what
HEADER: Test
HEADER: This works fine
Constructor called
HEADER: This doesn't use the copy-constructor
Constructor called
HEADER: This is a test
The code does exactly what I want but it depends on RVO. I read multiple times that you should not rely on RVO so I wanted to ask how I can:
Avoid RVO completely so that the copy constructor is called every time
Avoid the copy constructor
I already tried to avoid the copy constructor by returning a reference or moving but that segfaults. I guess thats because the temporary in Proxy::operator<< is deleted during return.
I'd also be interested in completely different approaches that do roughly the same.
This seems like a contrived problem: Firstly, the code works whether RVO is enabled or disabled (you can test it by using G++ with the no-elide-constructors flag). Secondly, the way you are designing the return of a Buffer object for use with the << operator can only be done by copying†: The Proxy::operator<<(const T& v) function creates a new Buffer instance on the stack, which is then deleted when you leave the function call (i.e. between each concatenation in pr << "This " << "doesn't " << "use the copy-constructor";); This is why you get a segmentation fault when trying to reference this object from outside the function.
Alternatively, you could define a << operator to use dynamic memory by e.g. returning a unique_ptr<Buffer>:
#include <memory>
...
std::unique_ptr<Buffer> operator<<(const T& v) const
{
if(v < 0)
return std::unique_ptr<Buffer>(new Buffer());
else
return std::unique_ptr<Buffer>(new Buffer(v));
}
However, your original concatenation statements won't be compilable, then, because Proxy::operator<<(const T& v) now returns an object of type std::unique_ptr<Buffer> rather than Buffer, meaning that this returned object doesn't have its own Proxy::operator<<(const T& v) function defined and so multiple concatenations won't work without first explicitly de-referencing the returned pointer:
const Proxy pr;
std::unique_ptr<Buffer> pb = pr << "This ";
// pb << "doesn't " << "use the copy-constructor"; // This line doesn't work
*pb << "doesn't " << "use the copy-constructor";
In other words, your classes rely inherently on copying and so, if you really want to avoid copying, you should throw them away and completely re-design your logging functionalities.
† I'm sure there's some black-magic voodoo which can be invoked to make this possible --- albeit at the cost of one's sanity.
My question is give below to avoid multiple copies in vector copying.
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
class DataValue {
public:
DataValue() { std::cout << "DataValue constructor called" << std::endl; }
DataValue(DataValue const& other) { cout << "DataValue copy constructor called" << std::endl; }
~DataValue() { std::cout << "DataValue destructor is called" << std::endl; }
private:
};
class ItemDataHistory {
public:
ItemDataHistory() { std::cout << "ItemDataHistory constructor called" << std::endl; }
// ItemDataHistory(ItemDataHistory const & other) { std::cout << "ItemDataHistory copy constructor called" << std::endl; }
~ItemDataHistory() { std::cout << "ItemDataHistory destructor called" << std::endl; }
std::vector<DataValue>& GetVecDataValues() { return m_vecDataValues; }
private:
std::vector<DataValue> m_vecDataValues;
};
class DataReply {
public:
DataReply() { std::cout << "Data reply constructor is called "<< std::endl; }
~DataReply() { std::cout << "Data reply destructor is called "<< std::endl; }
DataReply(const DataReply& ) { std::cout << "Data reply copy constructor is called "<< std::endl; }
std::vector<ItemDataHistory>& GetItemDataHistories() { return m_vecItemData; }
private:
// The list of DataValue
std::vector<ItemDataHistory> m_vecItemData;
};
void main()
{
DataValue dv1, dv2, dv3;
ItemDataHistory itmDH;
itmDH.GetVecDataValues().reserve(3);
itmDH.GetVecDataValues().push_back(dv1);
itmDH.GetVecDataValues().push_back(dv2);
itmDH.GetVecDataValues().push_back(dv3);
DataReply dr;
dr.GetItemDataHistories().reserve(1);
dr.GetItemDataHistories().push_back(itmDH); // Here copy consturtor of itemdatahistory is called and all data values are copied.
// Here I want to avoid data values constructor to be called again how can I avoid this
// How can I directly insert values of dv1, dv2, dv3 into "dr" with out using "itmDH"?
return;
}
Note here I cannot use pointer in above std::vector m_vecItemData; in data reply class as these are interface classes from libary and don't have control on it and I am calling function so function may use data while data in scope
My question is given in above comment in code. Reason is that I have thousands of data values. To avoid multiple constructors of data values to be called, I want to insert data values directly to data reply (i.e., with out using itmDH local variable)
and other questions is
How I can reserve space of data values inside data reply?
With C++11, you have two options:
make your type ItemDataHistory movable and move your data (if possible) with dr.GetItemDataHistories().push_back(std::move(itmDH));
look into new member function of containers, e.g. emplace_back().
In C++11, you can use move semantics.
Instead of doing this:
itmDH.GetVecDataValues().push_back(dv1);
itmDH.GetVecDataValues().push_back(dv2);
itmDH.GetVecDataValues().push_back(dv3);
You could do this:
itmDH.GetVecDataValues().push_back(std::move(dv1));
itmDH.GetVecDataValues().push_back(std::move(dv2));
itmDH.GetVecDataValues().push_back(std::move(dv3));
Instead of copying values, they are simply moved into the vector.
And instead of copying itmDH
dr.GetItemDataHistories().push_back(itmDH);
you could move it as well:
dr.GetItemDataHistories().push_back(std::move(itmDH));
In addition you also need move constructors. Here's an example:
DataValue(DataValue&& other){
std::cout << "DataValue move constructor called" << std::endl;
}
You may also declare and define move assignment operator:
DataValue& operator=(DataValue&& other){
std::cout << "DataValue move assigment operator is called" << std::endl;
return *this;
}
In order to fully understand move semantics (and rvalue references as well) please take a look at the following links:
http://www.cprogramming.com/c++11/rvalue-references-and-move-semantics-in-c++11.html
http://thbecker.net/articles/rvalue_references/section_01.html
I was going through Josuttis's "Using Map's as associative arrays" (from The C++ Standard Library - A Tutorial and Reference, 2nd Edition) and came across Using a std::map as an associative array on Stack Overflow. Now I have more questions on the constructors that are called when inserting into a map.
Here is my sample program (not using best coding practices; please excuse me for that):
class C
{
public:
string s;
C() { cout << "default " << endl;}
C(const string& p) : s(p)
{ cout << "one param" << endl;}
C(const C& obj)
{
if (this != &obj)
{
s = obj.s;
}
cout << "copy constr" << endl;
}
C& operator = (const C& obj)
{
if (this != &obj)
{
s = obj.s;
}
cout << "copy initializer" << endl;
return *this;
}
};
int main()
{
map<int,C> map1;
C obj("test");
cout << "Inserting using index" << endl;
map1[1] = obj;
cout << "Inserting using insert / pair" << endl;
map1.insert(make_pair(2,obj));
}
The output for this program is:
one param
Inserting using index
default
copy constr
copy constr
copy initializer
Inserting using insert / pair
copy constr
copy constr
copy constr
copy constr
I was assuming that initializing the map by index should call the default constructor and followed by the assignment operator.
But executing map1[1] = obj creates following output;
Inserting using index
default
copy constr
copy constr
copy initializer
Can someone help me to understand the initialization better?
If you read the specification for std::map, it says that operator[] is equivalent to (in this case)
(*((this->insert(make_pair(1,C()))).first)).second
So this explains all the constructor calls you see. First it calls the default constructor C(). Then it calls make_pair, which copies the C object. Then it calls insert, which makes a copy of the pair object you just made, calling the C copy constructor again. Finally it calls the assignment operator to set the inserted object to the one you are assigning it to.
Don;t know.
But this is interesting:
#include <string>
#include <map>
#include <iostream>
using namespace std;
class C
{
public:
string s;
C()
{
cout << "default " << endl;
}
C(const string& p)
: s(p)
{
cout << "one param(" << s << ")" << endl;
}
C(const C& obj)
:s(obj.s)
{
cout << "copy constr(" << s << ")" <<endl;
}
C& operator = (const C& obj)
{
cout << "copy initializer\t" <<;
C copy(obj);
std::swap(s,copy.s);
return *this;
}
};
int main()
{
map<int,C> map1;
cout << "Inserting using index" << endl;
map1[1] = C("Plop");
}
It looks like the default one is created and copied around.
Then the external one is just assinged over it once it has been put in place.
Inserting using index
default
copy constr()
copy constr()
one param(Plop)
copy initializer copy constr(Plop)
What happens if you simply execute map[1];? This may involve internal copies, depending on the implementation of map your standard library uses.
Actually map1[1] = obj will create pair first