Consider the following:
struct A { /* ... */ };
A foo() {
auto p = std::make_pair(A{}, 2);
// ... do something
return p.first;
}
auto a = foo();
Will p.first be copied, moved or RVO-ed?
I've found in Visual Studio 2010 and in gcc-5.1 RVO is not applied (see for example http://coliru.stacked-crooked.com/a/17666dd9e532da76).
The relevant section of the standard is 12.8.31.1 [class.copy]. It states that copy elision is permitted (my highlighting):
in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object (other than a function parameter or a variable introduced by the exception-declaration of a handler ([except.handle])) with the same type (ignoring cv-qualification) as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function's return value
Since p.first is not the name of an object, RVO is prohibited.
Just to add a little more fuel, how would this function if RVO were in play? The caller has put an instance of A somewhere in memory and then calls foo to assign to it (even better, let's assume that that A was a part of a larger struct, and let's assume that it is correctly aligned such that the next member of the struct is immediately after that instance of A). Assuming RVO were in play, the first portion of p is located where the caller wanted it, but where does the int that is second get placed? It has to go right after the instance of A in order to keep the pair functioning correctly, but at the source location, there's some other member right after that instance of A.
I would expect that RVO would not be happening in this place as you are only returning a portion of a larger object. A move could happen as first would have to be left in a destructible state.
#atkins got here first with the answer. Just adding this little test program which you may find useful in future when tracking move/assign behaviour.
#include <iostream>
#include <string>
using namespace std::string_literals;
struct A {
A()
: history("created")
{
}
A(A&& r)
: history("move-constructed,"s + r.history)
{
r.history = "zombie: was "s + r.history;
}
A(const A& r)
: history("copied from: " + r.history)
{
}
~A() {
history = "destroyed,"s + history;
std::cout << history << std::endl;
}
A& operator=(A&& r) {
history = "move-assigned from " + r.history + " (was "s + history + ")"s;
r.history = "zombie: was "s + r.history;
return *this;
}
A& operator=(const A&r ) {
history = "copied from " + r.history;
return *this;
}
std::string history;
};
A foo() {
auto p = std::make_pair(A{}, 2);
// ... do something
return p.first;
}
auto main() -> int
{
auto a = foo();
return 0;
}
example output:
destroyed,zombie: was created
destroyed,move-constructed,created
destroyed,copied from: move-constructed,created
Consider following code:
struct A {};
struct B {};
struct C { B c[100000]; };
A callee()
{
struct S
{
A a;
C c;
} s;
return s.a;
}
void caller()
{
A a = callee();
// here should lie free unused spacer of size B[100000]
B b;
}
"Partial" RVO should result in excessive stack usage bloating in caller, because (I think) S can be constructed only entirely in caller stack frame.
Another issue is ~S() behaviour:
// a.hpp
struct A {};
struct B {};
struct C { A a; B b; ~C(); };
// a.cpp
#include "a.hpp"
~C() { /* ... */; }
// main.cpp
#include "a.hpp"
A callee()
{
C c;
return c.a;
} // How to destruct c partially, having the user defined ~C() in another TU?
// Even if destructor is inline and its body is visible,
// how to automatically change its logic properly?
// It is impossible in general case.
void caller() { A a = callee(); }
Related
I would like to have a unique_ptr class member that points to the base class, but later in the constructor through polymorphism can be changed to point to a sister class that also derives from the same base class.
While I don't get any errors in the constructor setting this polymorphism, it does not seem to work correctly, since I get error messages that my polymorphic pointer can't find a member of the sister class to which I thought the pointer was now pointing.
How do I correctly achieve polymorphism here?
class A {
int bar;
};
class B : public A {
int foo;
};
class C: public A {
C();
std::unique_ptr<A> _ptr; // changing to std::unique_ptr<B> _ptr removes the "class A has no member 'foo'" error
};
C::C() : A()
{
_ptr = std::make_unique<B>(); // no errors here
int w = _ptr->foo; // class A has no member 'foo'
}
When you assign
_ptr = std::make_unique<B>();
This works because B is a derived class of A, however _ptr is still a unique_ptr to the base class. You can't change the type of a variable after it's declared.
So what are your options?
Because you know that _ptr stores a pointer to the derived class B, you can do a cast after dereferencing it:
_ptr = std::make_unique<B>();
// derefence the pointer, and cast the reference to `B&`.
B& reference_to_sister = (B&)(*_ptr);
int w = reference_to_sister.foo;
If you take this approach, you'll have to somehow keep track of which derived class is in _ptr, or you'll run the risk of running into bugs.
Alternatively, if you're using C++17, you can use std::variant:
class C : public A {
void initialize(A& a) {
// Do stuff if it's the base class
}
void initialize(B& b) {
// Do different stuff if it's derived
int w = b.foo;
}
C() {
_ptr = std::make_unique<B>(); // This works
// This takes the pointer, and calls 'initialize'
auto initialize_func = [&](auto& ptr) { initialize(*ptr); };
// This will call 'initialize(A&)' if it contains A,
// and it'll call 'initialize(B&)' if it contains B
std::visit(initialize_func, _ptr);
}
std::variant<std::unique_ptr<A>, std::unique_ptr<B>> _ptr;
};
In fact, if you use std::variant this will work even if A and B are completely unrelated classes.
Here's another short variant example
#include <variant>
#include <string>
#include <iostream>
void print(std::string& s) {
std::cout << "String: " << s << '\n';
}
void print(int i) {
std::cout << "Int: " << i << '\n';
}
void print_either(std::variant<std::string, int>& v) {
// This calls `print(std::string&) if v contained a string
// And it calls `print(int)` if v contained an int
std::visit([](auto& val) { print(val); }, v);
}
int main() {
// v is empty right now
std::variant<std::string, int> v;
// Put a string in v:
v = std::string("Hello, world");
print_either(v); //Prints "String: Hello, world"
// Put an int in v:
v = 13;
print_either(v); //Prints "Int: 13"
}
I've been paying close attention to the advice never to write std::move in a return statement, for example. Except there are some edge cases, for example.
I believe the following is another simple example of where std::move may be worthwhile - did I miss something? But I'm not sure why, and will that change in a future C++?
#include <iostream>
struct A
{
};
struct C
{
};
struct B
{
B(const A&, const C&) { std::cout << "B was copied\n"; }
B(A&&, C&&) { std::cout << "B was moved\n"; }
};
B f()
{
A a;
C c;
//return {a, c}; // Gives "B was copied"
return {std::move(a), std::move(c)}; // Gives "B was moved"
}
int main() {
f();
return 0;
}
return {std::move(a), std::move(c)}
is equivalent to
return B{std::move(a), std::move(c)}
You're basically invoking B::B(A&&, C&&) instead of the version taking const& references. This has nothing to do with moving a return value.
The return value of the function is the temporary instance of B, which is a prvalue. It C++17, it will benefit from "guaranteed copy elision". Before C++17, it will be RVOd or moved into its target.
Is there any way to create variable that will be unique for some lambda function and will last between launches of lambda?
More careful description: I want lambda with variable initialized to some value, and that variable should last between launches:
std::function<void(void)> a=[]()
{
/*here we declare variable X and initialize it to 0*/;
std::cout<<X++;
};
a();a();
So this should print out 01
But also I need to be sure that "X" is unique for "a", so after previous part this
std::function<void(void)> b=a;
b();b();
should print out 01.
I tried using static variables, but they are shared between copies(so these two parts print out 0123).
So, is there any way to do it?
I don't think mutable lambdas are sufficient. The mutable capture will get copied, when you copy the function pointer, also copying the counter. My read of the question, is that each copy of the lambda should start with the initial mutable capture.
You need to capture a custom class, with a copy constructor, to do this:
#include <functional>
#include <iostream>
class my_class {
public:
int n=0;
my_class()
{
}
my_class(const my_class &b)
{
}
};
int main()
{
std::function<void(void)> a=
[my_class_instance=my_class()]()
mutable
{
std::cout << my_class_instance.n++;
};
a();
a();
auto b=a;
b();
b();
}
The result from this is:
0101
Without a helper class, the equivalent code using only mutable lambdas will produce a
0123
My read of the question, is that the former behavior is desired.
You want it to reset on copies. Make data that does this:
template<class T>
struct no_copy {
T init;
T current;
operator T&(){ return current; }
operator T const&()const{ return current; }
no_copy( T&& i ):init(i), current(init) {}
no_copy( no_copy const&o ):init(o.init), current(init) {}
no_copy( no_copy &&o ):init(std::move(o.init)), current(init) {}
};
template<class T>
no_copy<std::decay_t<T>> make_no_copy(T&& t){
return {std::forward<T>(t)};
}
Then, in C++14, easy:
std::function<void(void)> a=[X=make_no_copy(0)]()mutable
{
std::cout<<X++;
};
a();a();
prints out 01.
In C++11:
auto X=make_no_copy(0);
std::function<void(void)> a=[X]()mutable
{
std::cout<<X++;
};
a();a();
it also works, but is a bit more ugly.
Other than a copy of X existing outside of the lambda, the C++11 version is the same as the C++14 version in behavior.
live example
Is using the copy constructor for "resetting" the only option? Shouldn't you be instead writing a factory function that emits fresh lambdas from the same initial environment?
Expecting that stuff A is different from stuff B after a copy is abuse of semantics.
auto make_counter() -> std::function<int()> {
return [x=0]() mutable { return x++; };
}
auto a = make_counter();
std::cout << a() << " " << a() << "\n";
auto b = make_counter();
std::cout << b() << " " << b() << "\n";
I wanted to restrict a specific class to be creatable on the stack only (not via allocation). The reason for this is that on the stack, the object which lifetime has begun last, will be the first to be destroyed, and I can create a hierarchy. I did it like this:
#include <cstddef>
#include <iostream>
class Foo {
public:
static Foo createOnStack() {
return {};
}
~Foo () {
std::cout << "Destructed " << --i << std::endl;
}
protected:
static int i;
Foo () {
std::cout << "Created " << i++ << std::endl;
}
Foo (const Foo &) = delete;
};
int Foo::i = 0;
The constructor normally should push the hierarchy stack, and the destructor pops it. I replaced it here for proof of concept. Now, the only way you can use such an object is by storing it in a temporary reference like this:
int main() {
Foo && a = Foo::createOnStack();
const Foo& b = Foo::createOnStack();
return 0;
}
My question now is, how safe is this with the C++ standard? Is there still a way to legally create a Foo on the heap or hand it down from your function into another frame (aka return it from your function) without running into undefined behaviour?
EDIT: link to example https://ideone.com/M0I1NI
Leaving aside the protected backdoor, C++17 copy elision breaks this in two ways:
#include<iostream>
#include<memory>
struct S {
static S make() {return {};}
S(const S&)=delete;
~S() {std::cout << '-' << this << std::endl;}
private:
S() {std::cout << '+' << this << std::endl;}
};
S reorder() {
S &&local=S::make();
return S::make();
}
int main() {
auto p=new S(S::make()),q=new S(S::make()); // #1
delete p; delete q;
reorder(); // #2
}
The use of new is obvious and has been discussed.
C++17 also allows prvalues to propagate through stack frames, which means that a local can get created before a return value and get destroyed while that return value is alive.
Note that the second case already existed (formally in C++14 and informally long before) in the case where local is of type S but the return value is some other (movable) type. You can't assume in general that even automatic object lifetimes nest properly.
I stumbled upon something similar today, and subsequently tried a few things out and noticed that the following seems to be legal in G++:
struct A {
int val_;
A() { }
A(int val) : val_(val) { }
const A& operator=(int val) { val_ = val; return *this; }
int get() { return val_; }
};
struct B : public A {
A getA() { return (((A)*this) = 20); } // legal?
};
int main() {
A a = 10;
B b;
A c = b.getA();
}
So B::getB returns a type A, after it as assigned the value 20 to itself (via the overloaded A::operator=).
After a few tests, it seems that it returns the correct value (c.get would return 20 as one may expect).
So I'm wondering, is this undefined behavior? If this is the case, what exactly makes it so? If not, what would be the advantages of such code?
After careful examination, with the help of #Kerrek SB and #Aaron McDaid, the following:
return (((A)*this) = 20);
...is like shorthand (yet obscure) syntax for:
A a(*this);
return a.operator=(20);
...or even better:
return A(*this) = 20;
...and is therefore defined behavior.
There are a number of quite separate things going on here. The code is valid, however you have made an incorrect assumption in your question. You said
"B::getA returns [...] , after it as assigned the value 20 to itself"
(my emphasis) This is not correct. getA does not modify the object. To verify this, you can simply place const in the method signature. I'll then fully explain.
A getA() const {
cout << this << " in getA() now" << endl;
return (((A)*this) = 20);
}
So what is going on here? Looking at my sample code (I've copied my transcript to the end of this answer):
A a = 10;
This declares an A with the constructor. Pretty straightfoward. This next line:
B b; b.val_ = 15;
B doesn't have any constructors, so I have to write directly to its val_ member (inherited from A).
Before we consider the next line, A c = b.getA();, we must very carefully consider the simpler expression:
b.getA();
This does not modify b, although it might superfically look like it does.
At the end, my sample code prints out the b.val_ and you see that it equals 15 still. It has not changed to 20. c.val_ has changed to 20 of course.
Look inside getA and you see (((A)*this) = 20). Let's break this down:
this // a pointer to the the variable 'b' in main(). It's of type B*
*this // a reference to 'b'. Of type B&
(A)*this // this copies into a new object of type A.
It's worth pausing here. If this was (A&)*this, or even *((A*)this), then it would be a simpler line. But it's (A)*this and therefore this creates a new object of type A and copies the relevant slice from b into it.
(Extra: You might ask how it can copy the slice in. We have a B& reference and we wish to create a new A. By default, the compiler creates a copy constructor A :: A (const A&). The compiler can use this because a reference B& can be naturally cast to a const A&.)
In particular this != &((A)*this). This might be a surprise to you. (Extra: On the other hand this == &((A&)*this) usually (depending on whether there are virtual methods))
Now that we have this new object, we can look at
((A)*this) = 20
This puts the number into this new value. This statement does not affect this->val_.
It would be an error to change getA such that it returned A&. First off, the return value of operator= is const A&, and therefore you can't return it as a A&. But even if you had const A& as the return type, this would be a reference to a temporary local variable created inside getA. It is undefined to return such things.
Finally, we can see that c will take this copy that is returned by value from getA
A c = b.getA();
That is why the current code, where getA returns the copy by value, is safe and well-defined.
== The full program ==
#include <iostream>
using namespace std;
struct A {
int val_;
A() { }
A(int val) : val_(val) { }
const A& operator=(int val) {
cout << this << " in operator= now" << endl; // prove the operator= happens on a different object (the copy)
val_ = val;
return *this;
}
int get() { return val_; }
};
struct B : public A {
A getA() const {
cout << this << " in getA() now" << endl; // the address of b
return (((A)*this) = 20);
// The preceding line does four things:
// 1. Take the current object, *this
// 2. Copy a slice of it into a new temporary object of type A
// 3. Assign 20 to this temporary copy
// 4. Return this by value
} // legal? Yes
};
int main() {
A a = 10;
B b; b.val_ = 15;
A c = b.getA();
cout << b.get() << endl; // expect 15
cout << c.get() << endl; // expect 20
B* b2 = &b;
A a2 = *b2;
cout << b2->get() << endl; // expect 15
cout << a2.get() << endl; // expect 15
}