It seems that initalizer lists are a good idea for your class constructors and, I'm assuming, for the copy constructor as well. For the assignment operator one has to assign each member in the body of the function. Consider the following simple block:
class Foo {
private:
int a,b;
public:
Foo(int c, int d) : a(c), b(d) {}
Foo(const Foo & X) : a(X.a), b(X.b) {}
Foo& operator=(const Foo& X) {
if (this == &X) return *this;
a = X.a;
b = X.b;
return *this;
}
};
If a class has a moderate amount of data members, there are three places where one can mess up the the different assignments/initialization. By that I mean, what if the copy constructor looked like:
Foo(const Foo & X) : a(X.a), b(X.a) {}
or a line was missing from the operator=. Since the assignment operator and the copy constructor often have the same effect (in that we copy members from one Foo to another) can I "reuse" the code from the copy constructor or the assignment operator or vice versa?
Your goal should be to not write copy constructors/assignment operators at all. Your goal should be to let the compiler do it. Standard library containers are all copyable, so use them where reasonable.
If there are members that cannot be copied correctly, then use smart pointers or other RAII objects. Those objects are the ones that should need special copy constructors/assignments. And they only need them for their one member.
Everything else should not use them.
Since the assignment operator and the copy constructor often have the same effect.
Not at all, one does initialization while the other does assignment. They are different in the initial state of the object, and their tasks are separate (though similar). The canonical assignment operator is usually done as:
Foo& operator=(Foo right) {
right.swap( *this );
return *this;
}
It might not be valid to forward all to an assignment operator, but that is was commmon in C++03 where it was allowed.
In C++11 constructors are easier: forward all the constructors to one master constructor.
class Foo {
private:
int a,b;
public:
Foo(int c, int d) : a(c), b(d) {}
Foo(const Foo & X) : Foo(x.a, x.d) {}
//syntax may be wrong, I don't have a C++11 compiler
Foo& operator=(const Foo& X) {
if (this == &X) return *this;
a = X.a;
b = X.b;
return *this;
}
}
In C++03 (where it is allowed)
class Foo {
private:
int a,b;
void init(int c, int d) {a=c; b=d;}
public:
Foo(int c, int d) : {init(c,d);}
Foo(const Foo & X) : {init(X.a, X.b);}
Foo& operator=(const Foo& X) { init(X.a, X.b);}
}
But keep in mind that this cannot be used in some common cases. (any object that isn't assignable)
Related
Let's say I have the following minimal example class:
#include <iostream>
class Foo {
public:
Foo() = default;
Foo(const Foo&) = default;
Foo(Foo&&) noexcept = default;
Foo& operator=(const Foo& rhs) {
std::cout << "copy\n";
return *this;
}
Foo& operator=(Foo&& rhs) noexcept {
std::cout << "move\n";
return *this;
}
Foo operator+(const Foo& rhs) const {
Foo x; // with some calculation
return x;
}
};
int main() {
Foo a, b, c;
a = b + c;
}
This prints move as expected. Now according to Effective C++ Item 3, I should return const Foo from operator+ to avoid construct like a + b = c, i.e.:
// To avoid a + b = c
const Foo operator+(const Foo& rhs) const {}
Unfortunately, this suddenly starts calling copy assignment instead of move assignment operator. [I'm using gcc 4.8.4 on Ubuntu, but it is probably nothing related to compiler]
How can I ensure that a + b = c fails to compile and in the same time move assignment is called for a = b + c? Or with the introduction of move semantics, is there no way to achieve both of them in the same time?
I have ended up using lvalue reference qualifier as pointed by Caninonos in comment and by max66 in now deleted answer (but 10k users can see it).
Foo& operator=(const Foo& rhs) & {}
Foo& operator=(Foo&& rhs) & noexcept {}
It is simple to implement and it provides a better interface design since assignment to anything other that lvalue doesn't sound meaningful and a possible source of bug.
However, it should be noted that the possibility of writing a + b = c by mistake is very low. Also compiler generated assignment operators are not lvalue reference qualified and we can write a + b = c with standard types, e.g. with std::string or with std::complex.
As title, can be to overloading operator = for casting?
I have a simple class.
class A{
protected:
int m_int;
public:
A& operator=( int& obj)
{
m_int = obj;
return *this;
}
};
I want:
A t_a = 1;
and
int t_int = t_a;
Is there a way to do this?
Just define conversion operator
operator int() const
{
return m_int;
}
or
explicit operator int() const
{
return m_int;
}
In the last case you have to use an explicit casting in the statement
int t_int = int( t_a );
Take into account that the assignment operator should be declared like
A& operator=( const int& obj)
{
m_int = obj;
return *this;
}
or like
A& operator=( int obj)
{
m_int = obj;
return *this;
}
Otherwise it will be impossible to bind the non-constant reference with integer literals or temporary values.
As for the assignment operator then you may define only a compound assignment operator for the type int and the type A.
For example you could define the operator += or something other operator.
Yes, that’s possible. You need a custom ctor and assignment operator. But writing those disables some of the compiler generated ctors/assignment ops. If you still need/want them, you need to reintroduce them explicitly.
class A
{
protected:
// Note the initializer. Without it m_int is uninitialized
// when you default construct an A. That’s a common source
// of bugs.
int m_int = 0;
public:
// Following two are the important ones
A(int i) : m_int(i) {}
A& operator=(int i)
{
m_int = i;
return *this;
}
// if A needs to be default constructible as well
A() = default;
// The int ctor/assignment disable the compiler generated
// normal copy ctor/assignment (the ones that take another A).
// Reintroduce them like this:
A(const A&) = default;
A& operator=(const A&) = default;
// Writing any copy ctor/assignment disables the compiler generated
// move ctor/assignment. If you still want them, reintroduce them.
A(A&&) = default;
A& operator=(A&&) = default;
};
A t_a = 1;
This doesn't use assignment. You need a constructor which takes an int argument.
int t_int = t_a;
You will need operator int() for this.
Note that it is a really bad idea to have a class which has both an implicit constructor from a type, and an implicit cast to the type. You will get all sorts of confusing errors when you try to do overload resolution.
Instead, I would make the constructor explicit, and write an explicit conversion function. That means you have to write:
int t_int = t_a.to_int();
But at least it's explicit.
Edit: Note that you can overload operator = for casting (either inside or outside the class), but neither of the code samples you gave will use it. = is used both for assignment and initialization, and both your samples are initialization (so won't use operator =)
It seems that adding a default constructor prevents from calling emplace_back and produces the error message: "static assertion failed: type is not assignable" (gcc 5.3 with -std=c++14). Here is a simple code that illustrates the issue:
class A {
public:
int a;
A() = default;
A(int a) {
this->a = a;
}
A(A const & a) = delete;
A& operator =(A const & a) = delete;
A(A && a) = default;
A& operator =(A && a) = default;
};
int main() {
A a(4);
std::vector<A> vec;
vec.emplace_back(std::move(a)); // Error: type is not assignable
return 0;
}
When removing the default constructor, the error goes away! Also, if the default constructor is defined (even if it does nothing), the error also goes away:
class A {
public:
int a;
A() {
}
A(int a) {
this->a = a;
}
A(A const & a) = delete;
A& operator =(A const & a) = delete;
A(A && a) = default;
A& operator =(A && a) = default;
};
int main() {
A b;
A a(4);
std::vector<A> vec;
vec.emplace_back(std::move(a)); // Error gone
return 0;
}
It seems that "A() = default;" is what is causing the problem.
Is this normal behaviour on part of the compiler or is it a bug?
It's a libstdc++ bug (edit: reported as bug 69478).
Briefly, libstdc++'s std::vector, as relevant here, uses std::uninitialized_copy (paired with move iterators) to move elements on reallocation, which is reduced to std::copy if the type is trivial and the iterators' reference types are assignable (i.e., the assignment operator that would conceptually be used is usable).
Then, the std::copy for pointers to trivial types (or in our case, a move_iterator wrapping a pointer) is in turn optimized into a call to memmove coupled with a check for is_copy_assignable. Of course, that check is wrong in this case, since the uninitialized_copy, paired with move iterators, only requires the thing to be move constructible.
When you don't have a default constructor or if the default constructor is user-defined, then the class isn't trivial, so you don't hit the code path that triggers this bug.
Why can I not check two objects of classes with explicit constructor only for equality? The following code does not compile
struct Foo
{
explicit Foo(int x) : x_(x) {}
int x_;
};
int main()
{
Foo(1) == Foo(1);
}
Do I have to declare operator == explicitly?
You need to overload the equality operator==:
struct Foo {
explicit Foo(int x) : x_(x) {}
int x_;
};
bool operator==(Foo const &lhs, Foo const& rhs) { return lhs.x_ == rhs.x_; }
LIVE DEMO
How shoud compiler know how it should compare them? Either define operator== or use Foo(1).x_ == Foo(1).x_ if you want to compare those ints.
Maybe you misunderstood what explicit constructor means. It's about asignment operator = and not comparism. Marking your constructor explicits disables following snippet to compile: Foo f = 1.
Yes, the compiler does not generate equality for you, so you have to do it yourself. This isn't about explicit constructors either; at no point has C++ ever allowed comparing classes or structs implicitly.
You need an operator == like that:
bool operator==(const Foo& f) { return x_==f.x_; }
I'm wondering if there is a way to implement copy constructors and assignment operators such that only a small modification is needed when these are redefined for a class.
For example, consider a class as such:
class Foo {
private:
int* mInt_ptr;
/* many other member variables
of different types that aren't
pointers */
public:
Foo();
Foo(const Foo&);
Foo& operator=(const Foo&);
~Foo();
};
Now, in order to deal with the pointer mInt_ptr I would need to handle it appropriately in the copy constructor and assignment operator. However, the rest of the member variables are safe to do a shallow copy of. Is there a way to do this automatically?
Once a class becomes large it may become tedious and unwieldy to explicitly write out the operations to copy the non-pointer member variables, so I'm wondering if there is a way to write, say, a copy constructor such as:
Foo::Foo(const Foo& tocopy)
{
mInt_ptr = new int(*tocopy.mInt_ptr);
/* Do shallow copy here somehow? */
}
rather than the explicit form of:
Foo::Foo(const Foo& tocopy)
{
mInt_ptr = new int(*tocopy.mInt_ptr);
mVar1 = tocopy.mVar1;
mVar2 = tocopy.mVar2;
...
...
mVarN = tocopy.mVarN;
}
Generally, don't use raw pointers, for exactly the reason that you're now fighting with. Instead, use a suitable smart pointer, and use copy-swap assignment:
class Foo
{
int a;
Zip z;
std::string name;
value_ptr<Bar> p;
public:
Foo(Foo const &) = default;
Foo & operator=(Foo rhs)
{
rhs.swap(*this);
return *this;
}
void swap(Foo & rhs)
{
using std::swap;
swap(a, rhs.a);
swap(z, rhs.z);
swap(name, rhs.name);
swap(p, rhs.p);
}
};
namespace std { template <> void swap<Foo>(Foo & a, Foo & b) { a.swap(b); } }
The value_ptr could be a full-blown implementation, or something simple such as this:
template <typename T> // suitable for small children,
class value_ptr // but not polymorphic base classes.
{
T * ptr;
public:
constexpr value_ptr() : ptr(nullptr) { }
value_ptr(T * p) noexcept : ptr(p) { }
value_ptr(value_ptr const & rhs) : ptr(::new T(*rhs.ptr)) { }
~value_ptr() { delete ptr; }
value_ptr & operator=(value_ptr rhs) { rhs.swap(*this); return *this; }
void swap(value_ptr & rhs) { std::swap(ptr, rhs.ptr); }
T & operator*() { return *ptr; }
T * operator->() { return ptr; }
};
How about you wrap all the shallow-copy bits in a small helper struct and use the default copy behaviour there.
class Foo {
private:
int* mInt_ptr;
struct helper_t
/* many other member variables
of different types that aren't
pointers */
} mHelper;
public:
Foo();
Foo(const Foo&);
Foo& operator=(const Foo&);
~Foo();
};
Foo::Foo(const Foo& tocopy)
{
mInt_ptr = new int(*tocopy.mInt_ptr);
mHelper = tocopy.mHelper;
}
Using better primitives, as Kerrek suggested, seems like better design though. This is just another possibility.
Regardless if you use raw pointers or smart pointers the Kerrek's solution is right in the sense that you should make a copy constructor, destructor and swap and implement assignment using those:
class Foo
{
private:
int* mInt_ptr;
// many other member variables
// of different types
public:
Foo()
: mInt_ptr(NULL)
// initialize all other members
{}
Foo(const Foo& that)
: mInt_ptr(new int(*that.mInt_ptr) )
// copy-construct all other members
{}
Foo& operator=(const Foo& that)
{
// you may check if(this == &that) here
Foo(that).swap(*this);
return *this;
}
~Foo()
{
delete mInt_ptr;
// and release other resources
}
void swap(Foo& that)
{
std::swap(mInt_ptr, that.mInt_ptr);
// swap all members
}
};
The members are inline here just to keep it compact, usually it is not advisable to burden class definition with inline member definitions.