Simple program:
#include <iostream>
using namespace::std;
class X {
public:
X() {
cout << "Default Constructor called\n";
i = 0;
}
X(int i) {
cout << "Parameterized Constructor called\n";
this->i = i;
}
X(const X& x) {
cout << "Copy Constructor called\n";
i = x.getI();
}
~X() {
cout << "Destructor called\n";
}
int getI() const {
return i;
}
X func() {
cout << "Entered func\n";
X x(2);
return x;
}
private:
int i;
};
int main() {
X x1;
X x2 = x1.func();
cout << "Returned from func\n";
}
It outputs the following:
Default Constructor called
Entered func
Parameterized Constructor called
Copy Constructor called
Destructor called
Returned from func
Destructor called
Destructor called
My issue is that after the 'Returned from func' is printed, no constructor is called when creating the instance x2. I was actually expecting a copy constructor to be called when instantiating x2 as it would have been if we did something like X x2 = x1;
I initially incorrectly assumed that the reason was RVO, so I keep my original answer below.
What your program prints is expected. Note that you print "Returned from function" after you call it and assign the returned value to x2. So the order of prints is correct: it first instantiates x, hence parametrized ctor call, then it copies it to x2, as it exits from the function, then it calls the destructor of x, which is no longer needed, then it actually exits from the function, and only then it reaches your print statement.
Old answer follows
What you observe is Return Value Optimization.
Return value optimization, or simply RVO, is a compiler optimization
technique that involves eliminating the temporary object created to
hold a function's return value.1 In C++, it is particularly notable
for being allowed to change the observable behaviour of the resulting
program.[2]
In general, the C++ standard allows a compiler to perform any
optimization, provided the resulting executable exhibits the same
observable behaviour as if (i.e. pretending) all the requirements of
the standard have been fulfilled. This is commonly referred to as the
"as-if rule".[3] The term return value optimization refers to a
special clause in the C++ standard which goes even further than the
"as-if" rule: an implementation may omit a copy operation resulting
from a return statement, even if the copy constructor has side
effects.[4]
See this wikipedia article, it has an example that is very similar to yours
Related
I am trying to understand what does the C++ standard say about how/when the destructor should be called when an object is returned from the function -
Consider this simple struct and two functions -
#include <iostream>
int g = 0;
struct foo {
int myid;
foo() {
myid = g;
g++;
std::cout << "Created " << myid << std::endl;
}
~foo() {
std::cout << "Destroyed " << myid << std::endl;
}
};
foo bar(void) {
int i = 0;
for (foo s; i < 10; i++) {
if (i == 5)
return s;
}
}
foo bar2(void) {
int i = 0;
foo s;
for (; i < 10; i++) {
if (i == 5)
return s;
}
}
int main() {
bar();
bar2();
return 0;
}
I am trying to track how many times the destructor is called. The output from the above program is -
Created 0
Destroyed 0
Destroyed 0
Created 1
Destroyed 1
I can understand the behavior of bar2. An object is created once and destroyed (I believe the destructor is called from main). But in bar when the object is declared inside the loop. It cases the destructor to be called twice. What is the reason for this discrepancy?
Is it the case that the standard leaves this behavior to the implementation (because of copy elision?) and g++ just choses this behavior for the two cases? If so how can I write this function so that I get predictable behavior. I need the destructor to be called the exact same number of times as the constructor (and preferably in the reverse order). I am okay with the destructor being called twice as long as the constructor is being called twice too. The reason is because I am allocating some data inside the constructor and freeing it inside the destructor.
Add this code
foo(const foo& rhs) {
myid = g;
g++;
std::cout << "Created from copy " << myid << std::endl;
}
This is a copy constructor, it's being called as well only you weren't aware of it, because you were using the default version, which obviously doesn't print anything, or increment your counter.
cppinsights tells you what's happening: There's a default copy constructor being called, so a copy is being destructed as well.
There, however, both objects are subject to named return value optimisation, a variant of copy elision that elides the copy constructor. If you compile and run your code with clang, that is indeed the case (https://godbolt.org/z/KWhRpL doesn't have the double "Destroyed").
NRVO is optional, and it seems like gcc doesn't apply it there. There is no way to force NRVO to happen, but you could implement a move constructor which will be called instead.
I'm working through chapter 18 of Stroustrup's Principles and Practice and am stuck on one part related to copy constructors.
I have a copy constructor defined as:
X(const X& x) {
out("X(X&)");
val = x.val;
}
X is a struct.
val is just an int value of X.
'out' is:
void out(const string& s) {
cerr << this << "->" << s << ": " << val << "\n";
}
I also have the following 2 functions defined:
X copy(X a) {
return a;
}
and
X copy2(X a) {
X aa = a;
return aa;
}
In main I have:
X loc(4);
X loc2 = loc;
loc2 = copy(loc);
loc2 = copy2(loc);
When I just call copy, the copy constructor is called twice: once for copy's parameter scope and once for the return call. This makes sense to me.
However, when I call copy2, the copy constructor is still just called twice: once for the function argument and once for 'X aa = a.' Why isn't it also called for the return?
There's no guarantee that copy constructors will be called in C++. In the case of return, it's likely to be replaced by a move or completely elided.
See also: What are copy elision and return value optimization?
Since you are returning a local variable, move semantics apply.
Optimizations make copying, moving and returning even more elaborate, see Tatuyuki Ishi's answer.
Here are some good examples for move semantics for return statements.
I'm currently trying to get a hang of move constructor.
I came upon the following (Compiled using g++ d.cpp --std=c++11 -O3)
class A {
string _x;
public:
A(string x) { cout << "default contrsutctor: " << x << "\n"; _x = x; }
A(const A& other) { cout << "copy contrsutctor: " << other._x << "\n"; _x = other._x; }
A(A&& other) { cout << "move contrsutctor: " << other._x << "\n"; _x = other._x; }
A foo() {
cout << "foo: " << _x << "\n";
return A("foo");
}
};
int main()
{
A a;
A b = a;
b.foo();
}
I expect this to output:
default contrsutctor: a
move contrsutctor: a
foo: a
default contrsutctor: foo
However the output is:
default contrsutctor: a
copy contrsutctor: a
foo: a
default contrsutctor: foo
Why isn't the A b = a line optimized to use the move constructor? The a object is never used afterwards, so it would be safe to optimize the code to use it instead of the copy constructor.
I know I could force the move contructor to be invoked with std::move(), but I'd prefer this to happen automatically in cases like this one.
Why isn't the A b = a line optimized to use the move constructor?
What you can do in copy constructor and move constructor could be totally different. The compiler cannot guarantee that the results of the two constructors are identical. Implementing this kind of optimization has the potential of changing the behavior of your program, which breaks the as-if rule.
You need to use std::move to cast a to A&&:
#include <utility>
int main()
{
A a("a");
A b = std::move(a);
b.foo();
}
A correct implementation of the move constructor should be:
A(A&& other)
: _x(std::move(other._x))
{}
After the line A b = std::move(a);, a should be "empty". In this case, a._x will be empty. as pointed by #TonyD in the comments, a._str could be in an unspecified but valid state (move constructor of std:string). You should use a with caution after this line.
A b = a; always invokes the copy constructor, no matter if it could invoke the move constructor. Additionally the lifetime of the object a continues after the assignment, even it is not used anymore.
If you want to use the move constructor, you have to make it explicit:
A b = std::move(a);
Note that this can be dangerous, as a is still accessible after the move. If you accidentally use it later, there may be undefined behavior.
Think about why it should happen automatically. In the example you gave, there is no need, as you can as well use a instead of b. In many cases where it would make more sense move constructor/assignment would be used automatically, e.g. A a; a = foo();.
Why isn't the A b = a line optimized to use the move constructor?
Because that would change the observable behavior of the program. The compiler is not permitted to freely change the observable behavior of the program (§1.9/1), except under very specific circumstances (§12.8/31). This is not one of those circumstances. Remove the side effects from your constructors, and the compiler may optimize them away. Of course, if you remove the side effects, then you won't notice if the compiler optimizes the constructor calls away (unless you examine the assembly or binary output), but that's the whole point.
I'm trying to establish whether it is safe for a C++ function to return an object that has a constructor and a destructor. My understanding of the standard is that it ought to be possible, but my tests with simple examples show that it can be problematic. For example the following program:
#include <iostream>
using namespace std;
struct My
{ My() { cout << "My constructor " << endl; }
~My() { cout << "My destructor " << endl; }
};
My function() { My my; cout << "My function" << endl; return my; }
int main()
{ My my = function();
return 0;
}
gives the output:
My constructor
My function
My destructor
My destructor
when compiled on MSVC++, but when compiled with gcc gives the following output:
My constructor
My function
My destructor
Is this a case of "undefined behavior", or is one of the compilers not behaving in a standard way? If the latter, which ? The gcc output is closer to what I would have expected.
To date, I have been designing my classes on the assumption that for each constructor call there will be at most one destructor call, but this example seems to show that this assumption does not always hold, and can be compiler-dependent. Is there anything in the standard that specifies what should happen here, or is it better to avoid having functions return non-trivial objects ? Apologies if this question is a duplicate.
In both cases, the compiler generates a copy constructor for you, that has no output so you won't know if it is called: See this question.
In the first case the compiler generated copy constructor is used, which matches the second destructor call. The line return my; calls the copy constructor, giving it the variable my to be used to construct the return value. This doesn't generate any output.
my is then destroyed. Once the function call has completed, the return value is destroyed at the end of the line { function();.
In the second case, the copy for the return is elided completely (the compiler is allowed to do this as an optimisation). You only ever have one My instance. (Yes, it is allowed to do this even though it changes the observable behaviour of your program!)
These are both ok. Although as a general rule, if you define your own constructor and destructor, you should also define your own copy constructor (and assignment operator, and possibly move constructor and move assignment if you have c++11).
Try adding your own copy constructor and see what you get. Something like
My (const My& otherMy) { cout << "My copy constructor\n"; }
The problem is that your class My violates the Rule of Three; if you write a custom destructor then you should also write a custom copy constructor (and copy assignment operator, but that's not relevant here).
With:
struct My
{ My() { cout << "My constructor " << endl; }
My(const My &) { cout << "My copy constructor " << endl; }
~My() { cout << "My destructor " << endl; }
};
the output for MSVC is:
My constructor
My function
My copy constructor
My destructor
My destructor
As you can see, (copy) constructors match with destructors correctly.
The output under gcc is unchanged, because gcc is performing copy elision as allowed (but not required) by the standard.
You are missing two things here: the copy constructor and NRVO.
The behavior seen with MSVC++ is the "normal" behavior; my is created and the rest of the function is run; then, when returning, a copy of your object is created. The local my object is destroyed, and the copy is returned to the caller, which just discards it, resulting in its destruction.
Why does it seem that you are missing a constructor call? Because the compiler automatically generated a copy constructor, which is called but doesn't print anything. If you added your own copy constructor:
My(const My& Right) { cout << "My copy constructor " << endl; }
you'd see
My constructor <----+
My function | this is the local "my" object
My copy constructor <--|--+
My destructor <----+ | this is the return value
My destructor <-----+
So the point is: it's not that there are more calls to destructors than constructors, it's just that you are not seeing the call to the copy constructor.
In the gcc output, you are also seeing NRVO applied.
NRVO (Named Return Value Optimization) is one of the few cases where the compiler is allowed to perform an optimization that alters the visible behavior of your program. In fact, the compiler is allowed to elide the copy to the temporary return value, and construct the returned object directly, thus eliding temporary copies.
So, no copy is created, and my is actually the same object that is returned.
My constructor <-- called at the beginning of f
My function
My destructor <-- called after f is terminated, since
the caller discarded the return value of f
To date, I have been designing my classes on the assumption that for each constructor call there will be at most one destructor call [...]
You can still "assume" that since it is true. Each constructor call will go in hand with exactly one destructor call. (Remember that if you handle stuff on the free/heap memory on your own.)
[..] and can be compiler-dependent [...]
In this case it can't. It is optimization depedant. Both, MSVC and GCC behave identically if optimization is applied.
Why don't you see identical behaviour?
1. You don't track everything that happens with your object. Compiler-generated functions bypass your output.
If you want to "follow-up" on the things your compiler does with your objects, you should define all of the special members so you can really track everything and do not get bypassed by any implicit function.
struct My
{
My() { cout << "My constructor " << endl; }
My(My const&) { cout << "My copy-constructor " << endl; }
My(My &&) { cout << "My move-constructor " << endl; }
My& operator=(My const&) { cout << "My copy-assignment " << endl; }
My& operator=(My &&) { cout << "My move-assignment " << endl; }
~My() { cout << "My destructor " << endl; }
};
[Note: The move-constructor and move-assignment will not be implicitly present if you have the copy ones but it's still nice to see when the compiler use which of them.]
2. You don't compile with optimization on both MSVC and GCC.
If compiled with MSVC++11 /O2 option the output is:
My constructor
My function
My destructor
If compiled in debug mode / without optimization:
My constructor
My function
My move-constructor
My destructor
My destructor
I can't do a test on gcc to verify if there's an option that enforces all of these steps but -O0 should do the trick I guess.
What's the difference between optimized and non-optimized compilation here?
The case without any copy omittance:
The completely "non-optimized" behaviour in this line My my_in_main = function();
(changed the name to make things clear) would be:
Call function()
In function construct My My my;
Output stuff.
Copy-construct my into the return value instance.
return and destroy my instance.
Copy(or move in my example)-construct the return value instance into my_in_main.
Destroy the return value instance.
As you can see: we have at most two copies (or one copy and one move) here but the compilers may possibly omit them.
To my understanding, the first copy is omited even without optimization turned on (in this case), leaving the process as follows:
Call function()
In function construct My My my; First constructor output!
Output stuff. Function output!
Copy(or move in my example)-construct the return value instance into my_in_main. Move output!
Destroy the return value instance. Destroy output!
The my_in_main is destroy at the end of main giving the last Destroy output!. So we know what happens in the non-optimized case now.
Copy elision
The copy (or move if the class has a move constructor as in my example) can be elided.
§ 12.8 [class.copy] / 31
When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the copy/move constructor and/or destructor for the object have side effects.
So now the question is when does this happen in this example? The reason for the elison of the first copy is given in the very same paragraph:
[...] 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 or catch-clause parameter) with the same cvunqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value.
Return type matches type in the return statement: function will construct My my; directly into the return value.
The reason for the elison of the second copy/move:
[...] when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move.
Target type matches the type returned by the function: The return value of the function will be constructed into my_in_main.
So you have a cascade here:
My my; in your function is directly constructed into the return value which is directly constructed into my_in_main So you have in fact only one object here and function() would (whatever it does) in fact operate on the object my_in_main.
Call function()
In function construct My instance into my_in_main. Constructor output!
Output stuff. Function output!
my_in_main is still destroyed at the end of main giving a Destructor output!.
That makes three outputs in total: Those you observe if optimization is turned on.
An example where elision is not possible.
In the following example both copies mentioned above cannot be omitted because the class types do not match:
The return statement does not match the return type
The target type does not match the return type of the function
I just created two additional types:
#include <iostream>
using namespace std;
struct A
{
A(void) { cout << "A constructor " << endl; }
~A(void) { cout << "A destructor " << endl; }
};
struct B
{
B(A const&) { cout << "B copy from A" << endl; }
~B(void) { cout << "B destructor " << endl; }
};
struct C
{
C(B const &) { cout << "C copy from B" << endl; }
~C(void) { cout << "C destructor " << endl; }
};
B function() { A my; cout << "function" << endl; return my; }
int main()
{
C my_in_main(function());
return 0;
}
Here we have the "completely non-optimized behaviour" I mentioned above. I'll refer to the points I've drawn there.
A constructor (see 2.)
function (see 3.)
B copy from A (see 4.)
A destructor (see 5.)
C copy from B (see 6.)
B destructor (see 7.)
C destructor (instance in main, destroy at end of main)
consider the following:
class X {
public:
X(int i) { cout << "X(int i)" << endl; }
X(const X& x) { cout << "X(const X& x)" << endl; }
};
void main() {
X x1(1);
X x2 = X(1);
X x3 = (X)1;
}
running this code produces this output:
X(int i)
X(int i)
X(int i)
I thought that all of the above three statements are equivalent as the copy CTOR is never called. However, changing X's copy CTOR to be private:
class X {
public:
X(int i) { cout << "X(int i)" << endl; }
private:
X(const X& x) { cout << "X(const X& x)" << endl; }
};
Will fail to compile (In visual studio 2010) with this error:
cannot access private member declared in class 'X'
So it seems the copy CTOR is involved somehow though I don't quite understand how.
Thanks
X x1(1);
X x2 = X(1);
X x3 = (X)1;
The reason is that all of these are not exactly equivalent.
First one is direct-initialization, while the second and third is copy-initialization. For copy-initialization, copy-constructor must be public, or else compiler will give error.
Now the question is, if 2nd and 3rd requires the copy-ctor to be public,then why the following output:
X(int i)
X(int i)
X(int i)
This surely says that copy-ctor is never called which is true. Compiler just elided the call to copy-ctor. According to §8.5/14, in such cases, the compiler is permitted to eliminate the need to call copy-constructor. That is why you don't see copy-ctor being called.
A little inside : in the 2nd and 3rd case, first a temporary is created by calling X(int i), then this temporary was supposed to be passed to the copy-ctor to copy-initialize the object being declared. But the compiler optimizes away this step, eliding the call to copy-ctor.
The X x2 = ... invokes the copy constructor (even if the compiler optimises it out later). Thus, it must still be accessible.
This :
X x3 = (X)1;
is c-style cast from int into the object of type X
This copying :
X x2 = X(1);
is optimized away but the compiler still needs the access to the copy-constructor.
Ist object makes use of parrametrized constructor as you know it and all others are using copy constructor. i.e. why when you are makin it private access violation occurs. The others two objects are making use of copy constructor.
These are the way of using copy constructor.