Different performance for default initialized and value initialized struct - c++

I have a simple struct with an array:
struct A
{
uint32_t arr[size];
};
I have two functions, which create it using default initialization and value initialization:
template<class T>
void testDefault()
{
T* pa = new T; // Default
use(*pa);
delete pa;
}
template<class T>
void testValue()
{
T* pa = new T(); // Value
use(*pa);
delete pa;
}
I'm facing different performance for those functions. The funny thing is that performance differences vary depending on how I declare default constructor of the struct. I have three ways:
struct A
{
uint32_t arr[size];
// Implicit constructor
};
struct B
{
uint32_t arr[size];
B() {}; // Empty constructor
};
struct C
{
uint32_t arr[size];
C() = default; // Defaulted constructor
};
I thought they are all the same from compiler's point of view. Never have been I so wrong. I did run both testDefault() and testValue() several times with structs A, B and C and measured performance. Here is what I have:
Default initialization (implict constructor) done in 880ms
Value initialization (implict constructor) done in 1145ms
Default initialization (empty constructor) done in 867ms
Value initialization (empty constructor) done in 865ms
Default initialization (defaulted constructor) done in 872ms
Value initialization (defaulted constructor) done in 1148ms
Note how performance is clearly worse for both implicit and defaulted constructors. Only empty constructor correctly shows the same performance for both different initialization forms.
I tested this with VC++, gcc and clang. See online demo for gcc. Timings are quite persistent.
What I assume is:
Default and value initializations for UDT are the same thing
All demonstrated ways of defining default constructor are doing the same thing
Default constructor of these structs should leave content of the array in indeterminate state
Since all the compilers exhibit the same timings, it seems like I'm missing something. Can anyone please explain me these timings?
(See also my question Why compilers put zeros into arrays while they do not have to? on the same topic. I give some links to cppreference there.)

Let's look at the definition of value-initialize:
To value-initialize an object of type T means:
if T is a (possibly cv-qualified) class type with either no default constructor (12.1) or a default constructor that is user-provided or deleted, then the object is default-initialized;
if T is a (possibly cv-qualified) class type without a user-provided or deleted default constructor, then the object is zero-initialized [...];
if T is an array type, then each element is value-initialized;
otherwise, the object is zero-initialized.
Also let's review the adjectives in use here for constructors:
user-declared - you declared the constructor
user-provided - you declared the constructor and didn't set it to = default or = delete
default - can be called with no arguments
declared as defaulted - marked = default; or implicitly generated as such
Looking at your classes:
A has a default constructor which is implicitly declared as defaulted, and not user-provided.
C has a default constructor which is user-declared as defaulted, and not user-provided.
So the second bullet point in the definition of value-initialize applies. The object is zero-initialized, meaning the arr is zeroed out.
B has a default constructor which is user-provided.
So the first bullet point of value-initialize applies to B; value-initialization is the same as default-initialization here, and arr is not zeroed out.
Your timings correctly seem to correspond to what is expected: value-initialization of A or C zeroes out arr and the other cases don't.

One of the constructors behaves differently under value initialization. In this version,
B() {};
the array B::arr is not value-initialized when the B is. With the others, it is. Whether this explains the performance difference is another matter.
So, B::arr doesn't get zero-initialized with value initialization, whereas A::arr and C::arr do. All three cases have the same behaviour under default initialization, that is to say, arr gets default-initialized, i.e. no initialization is performed.

Related

C++ : Default constructor for struct doesn't work? [duplicate]

The C++ standard (section 8.5) says:
If a program calls for the default initialization of an object of a const-qualified type T, T shall be a class type with a user-provided default constructor.
Why? I can't think of any reason why a user-provided constructor is required in this case.
struct B{
B():x(42){}
int doSomeStuff() const{return x;}
int x;
};
struct A{
A(){}//other than "because the standard says so", why is this line required?
B b;//not required for this example, just to illustrate
//how this situation isn't totally useless
};
int main(){
const A a;
}
The reason is that if the class doesn't have a user-defined constructor, then it can be POD, and the POD class is not initialized by default. So if you declare a const object of POD which is uninitialized, what use of it? So I think the Standard enforces this rule so that the object can actually be useful.
struct POD
{
int i;
};
POD p1; //uninitialized - but don't worry we can assign some value later on!
p1.i = 10; //assign some value later on!
POD p2 = POD(); //initialized
const POD p3 = POD(); //initialized
const POD p4; //uninitialized - error - as we cannot change it later on!
But if you make the class a non-POD:
struct nonPOD_A
{
nonPOD_A() {} //this makes non-POD
};
nonPOD_A a1; //initialized
const nonPOD_A a2; //initialized
Note the difference between POD and non-POD.
User-defined constructor is one way to make the class non-POD. There are several ways you can do that.
struct nonPOD_B
{
virtual void f() {} //virtual function make it non-POD
};
nonPOD_B b1; //initialized
const nonPOD_B b2; //initialized
Notice nonPOD_B doesn't defined user-defined constructor. Compile it. It will compile:
http://www.ideone.com/h7TsA
And comment the virtual function, then it gives error, as expected:
http://www.ideone.com/SWk7B
Well, I think, you misunderstood the passage. It first says this (§8.5/9):
If no initializer is specified for an object, and the object is of (possibly cv-qualified) non-POD class type (or array thereof), the object shall be default-initialized; [...]
It talks about non-POD class possibly cv-qualified type. That is, the non-POD object shall be default-initialized if there is no initializer specified. And what is default-initialized? For non-POD, the spec says (§8.5/5),
To default-initialize an object of type T means:
— if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
It simply talks about default constructor of T, whether its user-defined or compiler-generated is irrelevant.
If you're clear up to this, then understand what the spec next says ((§8.5/9),
[...]; if the object is of const-qualified type, the underlying class type shall have a user-declared default constructor.
So this text implies, the program will be ill-formed if the object is of const-qualified POD type, and there is no initializer specified (because POD are not default initialized):
POD p1; //uninitialized - can be useful - hence allowed
const POD p2; //uninitialized - never useful - hence not allowed - error
By the way, this compiles fine, because its non-POD, and can be default-initialized.
Pure speculation on my part, but consider that other types have a similar restriction, too:
int main()
{
const int i; // invalid
}
So not only is this rule consistent, but it also (recursively) prevents unitialized const (sub)objects:
struct X {
int j;
};
struct A {
int i;
X x;
}
int main()
{
const A a; // a.i and a.x.j in unitialized states!
}
As for the other side of the question (allowing it for types with a default constructor), I think the idea is that a type with a user-provided default constructor is supposed to always be in some sensible state after construction. Note that the rules as they are allow for the following:
struct A {
explicit
A(int i): initialized(true), i(i) {} // valued constructor
A(): initialized(false) {}
bool initialized;
int i;
};
const A a; // class invariant set up for the object
// yet we didn't pay the cost of initializing a.i
Then perhaps we could formulate a rule like 'at least one member must be sensibly initialized in a user-provided default constructor', but that's way too much time spent trying to protect against Murphy. C++ tends to trust the programmer on certain points.
This was considered a defect (against all versions of the standard) and it was resolved by Core Working Group (CWG) Defect 253. The new wording for the standard states in http://eel.is/c++draft/dcl.init#7
A class type T is const-default-constructible if
default-initialization of T would invoke a user-provided constructor
of T (not inherited from a base class) or if
each direct non-variant non-static data member M of T has a default member initializer or, if M is of class type X (or array thereof), X
is const-default-constructible,
if T is a union with at least one non-static data member, exactly one variant member has a default member initializer,
if T is not a union, for each anonymous union member with at least one non-static data member (if any), exactly one non-static data
member has a default member initializer, and
each potentially constructed base class of T is const-default-constructible.
If a program calls for the default-initialization of an object of a
const-qualified type T, T shall be a const-default-constructible class
type or array thereof.
This wording essentially means that the obvious code works. If you initialize all of your bases and members, you can say A const a; regardless of how or if you spell any constructors.
struct A {
};
A const a;
gcc has accepted this since 4.6.4. clang has accepted this since 3.9.0. Visual Studio also accepts this (at least in 2017, not sure if sooner).
I was watching Timur Doumler's talk at Meeting C++ 2018 and I finally realised why the standard requires a user-provided constructor here, not merely a user-declared one. It has to do with the rules for value initialisation.
Consider two classes: A has a user-declared constructor, B has a user-provided constructor:
struct A {
int x;
A() = default;
};
struct B {
int x;
B() {}
};
At first glance, you might think these two constructors will behave the same. But see how value initialisation behaves differently, while only default initialisation behaves the same:
A a; is default initialisation: the member int x is uninitialised.
B b; is default initialisation: the member int x is uninitialised.
A a{}; is value initialisation: the member int x is zero-initialised.
B b{}; is value initialisation: the member int x is uninitialised.
Now see what happens when we add const:
const A a; is default initialisation: this is ill-formed due to the rule quoted in the question.
const B b; is default initialisation: the member int x is uninitialised.
const A a{}; is value initialisation: the member int x is zero-initialised.
const B b{}; is value initialisation: the member int x is uninitialised.
An uninitialised const scalar (e.g. the int x member) would be useless: writing to it is ill-formed (because it's const) and reading from it is UB (because it holds an indeterminate value). So this rule prevents you from creating such a thing, by forcing you to either add an initialiser or opt-in to the dangerous behaviour by adding a user-provided constructor.
I think it would be nice to have an attribute like [[uninitialized]] to tell the compiler when you're intentionally not initialising an object. Then we wouldn't be forced to make our class not trivially default constructible to get around this corner case. This attribute has actually been proposed, but just like all the other standard attributes, it does not mandate any normative behaviour, being merely a hint to the compiler.
Congratulations, you've invented a case in which there need not be any user defined constructor for the const declaration with no initializer to make sense.
Now can you come up with a reasonable re-wording of the rule that covers your case but still makes the cases that should be illegal illegal? Is it less than 5 or 6 paragraphs? Is it easy and obvious how it should be applied in any situation?
I posit that coming up with a rule that allows the declaration you created to make sense is really hard, and making sure that the rule can be applied in a way that makes sense to people when reading code is even harder. I would prefer a somewhat restrictive rule that was the right thing to do in most cases to a very nuanced and complex rule that was difficult to understand and apply.
The question is, is there a compelling reason the rule should be more complex? Is there some code that would otherwise be very difficult to write or understand that can be written much more simply if the rule is more complex?

What Form Does the Implicitly-Declared Default Constructor Take?

So let's say I'm working with this toy example:
struct Foo {
int member;
};
I know that the default constructor won't default initialize member. So if I do this, member remains uninitialized: const Foo implicit_construction. As an aside this seems to work fine: const Foo value_initialization = Foo() though my understanding is that this isn't actually using the default constructor.
If I change Foo like this:
struct Foo {
Foo() = default;
int member;
};
And I try to do const Foo defaulted_construction, unsurprisingly it behaves exactly as implicit_construction did, with member being uninitialized.
Finally, if I change Foo to this:
struct Foo {
Foo(){};
int member;
};
And I do: const Foo defined_construction member is zero-initialized. I'm just trying to make sense of what the implicitly defined constructor looks like. I would have though it would have been Foo(){}. Is this not the case? Is there some other black magic at work that makes my defined constructor behave differently than the defaulted one?
Edit:
Perhaps I'm being mislead here. defaulted_construction is definitely uninitialized.
While defined_construction is definitely initialized.
I took this to be standardized behavior, is that incorrect?
What you're experiencing is called default initialization and the rules for it are (emphasis mine):
if T is a non-POD (until C++11) class type, the constructors are considered and subjected to overload resolution against the empty argument list. The constructor selected (which is one of the default constructors) is called to provide the initial value for the new object;
if T is an array type, every element of the array is default-initialized;
otherwise, nothing is done: the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.
Edit, in response to OP's request below:
Note that declaring a constructor = default does not change the situation (again, emphasis mine):
Implicitly-defined default constructor
If the implicitly-declared default constructor is not defined as deleted, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used, and it has exactly the same effect as a user-defined constructor with empty body and empty initializer list. That is, it calls the default constructors of the bases and of the non-static members of this class.
Since the default constructor has an empty initializer list, its members satisfy the conditions for default initialization:
Default initialization is performed in three situations:
...
3) when a base class or a non-static data member is not mentioned in a constructor initializer list and that constructor is called.
Also note that you have to be careful when experimentally confirming this, because it's entirely possible that the default-initialized value of an int may be zero. In particular, you mentioned that this:
struct Foo {
Foo(){};
int member;
} foo;
Results in value-initialization, but it does not; here, member is default-initialized.
Edit 2:
Note the following distinction:
struct Foo {
int member;
};
Foo a; // is not value-initialized; value of `member` is undefined
Foo b = Foo(); // IS value-initialized; value of `member` is 0
This behavior can be understood by following the rules for value-initialization:
Value initialization is performed in these situations:
1,5) when a nameless temporary object is created with the initializer consisting of an empty pair of parentheses;
Form 1 (T();) is the form used on the right-hand side of the = above to initialize b.
The effects of value initialization are:
1) if T is a class type with no default constructor or with a user-provided or deleted default constructor, the object is default-initialized;
2) if T is a class type with a default constructor that is neither user-provided nor deleted (that is, it may be a class with an implicitly-defined or defaulted default constructor), the object is zero-initialized and then it is default-initialized if it has a non-trivial default constructor;
3) if T is an array type, each element of the array is value-initialized;
4) otherwise, the object is zero-initialized.
Finally though, note that in our earlier example:
struct Foo {
Foo(){}; // this satisfies condition (1) above
int member;
};
Foo f = Foo();
Now, condition (1) applies, and our (empty) user-declared constructor is called instead. Since this constructor does not initialize member, member is default-initialized (and its initial value is thus undefined).

C++ direct list initialization with deleted default constructor

I have the following class definition. I included a private x to make sure it is not an aggregate.
class A {
private:
int x;
public:
A() = delete;
A(std::initializer_list<int>) { printf("init\n"); }
};
Now if I initialize this object with A a = A{}, it will say A::A() is deleted. I guess it is trying to call A::A() but it is deleted. If I comment out that line, so A::A() is automatically generated. Then if I run this code, I could see that it is calling A::A(std::initializer_list<int>)!
And more confusing, if I define A() = default, the initialization calls A::A() again.
Can anyone point me to the right direction of understading this behavior? Thanks!
I'm using G++ 6.3.0 with flag -std=c++17.
The behaviour is correct.
First of all:
// Calling
A a = A{}; // equals A a = A();,
which was a convenience uniformization of the list-initialization idiom.
Refer to: https://stackoverflow.com/a/9020606/3754223
Case A:
class A {
private:
int x;
public:
**A() = delete;**
A(std::initializer_list<int>) { printf("init\n"); }
};
As said above, A a = A{} will be ... = A(), which is deleted.
Case B:
class A {
private:
int x;
public:
A(std::initializer_list<int>) { printf("init\n"); }
};
In this case, there's no default constructor provided, since you have your initializer-list constructor be defined. Consequently {} is converted to an empty initializer list implicitely!
Case C:
class A {
private:
int x;
public:
**A() = default;**
A(std::initializer_list<int>) { printf("init\n"); }
};
In this case the empty default constructor is available and A{} is converted back to A(), causing it to be called.
Please refer to the below pages and get a thorough read into it.
http://en.cppreference.com/w/cpp/utility/initializer_list
http://en.cppreference.com/w/cpp/language/value_initialization
Finally:
Using:
A a = {{}};
Would cause that ALWAYS the initializer-list constructor is called, as the outer-curly-brackets denoted the init-list, and the inner a zero-constructed element. -> Non-empty initializer-list... (See stackoverflow link above again!)
And btw, I cannot understand why the question is downvoted, considering that this is a really tricky part...
You have lots of cases listed, so let us go over them one by one.
A() = delete;
A(std::initializer_list<int>) { ... }
Writing A a = A{}; will indeed try to call the deleted default constructor, and your code fails to compile.
What you have above is list initialization, and because the braced-init-list is empty, value initialization will be performed. This selects the deleted default constructor, which makes your code ill-formed.
If I comment out that line, so A::A() is automatically generated
No, this is not the case. There is no implicitly declared default constructor due to the presence of the user provided constructor that takes an initializer_list. Now, A a = A{}; will call the initializer_list constructor, and in this case the initializer_list will be empty.
if I define A() = default, the initialization calls A::A() again
This behaves the same as the first case, except the default constructor is explicitly defaulted instead of deleted, so your code compiles.
Finally,
I included a private x to make sure it is not an aggregate
There's no need for this, defining a constructor makes A a non-aggregate.
List initialization follows a very specific ordering, as laid out in [dcl.init.list]/3. The two relevant bullet points and their relative ordering are highlighted:
List-initialization of an object or reference of type T is defined as follows:
— If T is an aggregate class and [...]
— Otherwise, if T is a character array and [...]
— Otherwise, if T is an aggregate, [...]
— Otherwise, if the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized.
— Otherwise, if T is a specialization of std::initializer_list<E>, [...]
— Otherwise, if T is a class type, constructors are considered. The applicable constructors are enumerated and the best one is chosen through overload resolution (13.3, 13.3.1.7).
— Otherwise, if the initializer list has a single element of type E and [...]
— Otherwise, if T is a reference type, [...]
— Otherwise, if T is an enumeration with a fixed underlying type (7.2), [...]
— Otherwise, if the initializer list has no elements, the object is value-initialized.
— Otherwise, the program is ill-formed.
An empty initializer list for class types with default constructors is a special case that precedes normal constructor resolution. Your class isn't an aggregate (since it has a user-provided constructor), so A{} will attempt to value-initialize A if A has a default constructor. It doesn't matter if that default constructor is accessible, it only matters if it exists. If it exists and is inaccessible (your first case), that's ill-formed. If it exists and is accessible (your third case), then it's invoked. Only if your class does not have a default constructor will the constructors be enumerated, in which case the A(initializer_list<int> ) constructor will be invoked with an empty initializer list (your second case).

Move constructor is required even if it is not used. Why?

Why?! Why C++ requires the class to be movable even if it's not used!
For example:
#include <iostream>
using namespace std;
struct A {
const int idx;
// It could not be compileld if I comment out the next line and uncomment
// the line after the next but the moving constructor is NOT called anyway!
A(A&& a) : idx(a.idx) { cout<<"Moving constructor with idx="<<idx<<endl; }
// A(A&& a) = delete;
A(const int i) : idx(i) { cout<<"Constructor with idx="<<i<<endl; }
~A() { cout<<"Destructor with idx="<<idx<<endl; }
};
int main()
{
A a[2] = { 0, 1 };
return 0;
}
The output is (the move constructor is not called!):
Constructor with idx=0
Constructor with idx=1
Destructor with idx=1
Destructor with idx=0
The code can not be compiled if moving constructor is deleted ('use of deleted function ‘A::A(A&&)’'. But if the constructor is not deleted it is not used!
What a stupid restriction?
Note:
Why do I need it for? The practical meaning appears when I am trying to initialize an array of objects contains unique_ptr field.
For example:
// The array of this class can not be initialized!
class B {
unique_ptr<int> ref;
public:
B(int* ptr) : ref(ptr)
{ }
}
// The next class can not be even compiled!
class C {
B arrayOfB[2] = { NULL, NULL };
}
And it gets even worse if you are trying to use a vector of unique_ptr's.
Okay. Thanks a lot to everybody. There is big confusion with all those copying/moving constructors and array initialization. Actually the question was about the situation when the compiler requires a copying consrtuctor, may use a moving construcor and uses none of them.
So I'm going to create a new question a little bit later when I get a normal keyboard. I'll provide the link here.
P.S. I've created more specific and clear question - welcome to discuss it!
A a[2] = { 0, 1 };
Conceptually, this creates two temporary A objects, A(0) and A(1), and moves or copies them to initialise the array a; so a move or copy constructor is required.
As an optimisation, the move or copy is allowed to be elided, which is why your program doesn't appear to use the move constructor. But there must still be a suitable constructor, even if its use is elided.
A a[2] = { 0, 1 };
This is aggregate initialization. §8.5.1 [dcl.init.aggr]/p2 of the standard provides that
When an aggregate is initialized by an initializer list, as specified in 8.5.4, the elements of the initializer list are taken as initializers for the members of the aggregate, in increasing subscript or member order. Each member is copy-initialized from the corresponding initializer-clause.
§8.5 [dcl.init]/p16, in turn, describes the semantics of copy initialization of class type objects:
[...]
If the destination type is a (possibly cv-qualified) class type:
If the initialization is direct-initialization, or if it is copy-initialization where the cv-unqualified version of the source
type is the same class as, or a derived class of, the class of the
destination, constructors are considered. The applicable constructors
are enumerated (13.3.1.3), and the best one is chosen through overload
resolution (13.3). The constructor so selected is called to initialize
the object, with the initializer expression or expression-list as its
argument(s). If no constructor applies, or the overload resolution is
ambiguous, the initialization is ill-formed.
Otherwise (i.e., for the remaining copy-initialization cases), user-defined conversion sequences that can convert from the source
type to the destination type or (when a conversion function is used)
to a derived class thereof are enumerated as described in 13.3.1.4,
and the best one is chosen through overload resolution (13.3). If the
conversion cannot be done or is ambiguous, the initialization is
ill-formed. The function selected is called with the initializer
expression as its argument; if the function is a constructor, the call
initializes a temporary of the cv-unqualified version of the
destination type. The temporary is a prvalue. The result of the call
(which is the temporary for the constructor case) is then used to
direct-initialize, according to the rules above, the object that is
the destination of the copy-initialization. In certain cases, an
implementation is permitted to eliminate the copying inherent in this
direct-initialization by constructing the intermediate result directly
into the object being initialized; see 12.2, 12.8.
Since 0 and 1 are ints, not As, the copy initialization here falls under the second clause. The compiler find a user-defined conversion from int to A in your A::A(int) constructor, calls it to construct a temporary of type A, then performs direct initialization using that temporary. This, in turn, means the compiler is required to perform overload resolution for a constructor taking a temporary of type A, which selects your deleted move constructor, which renders the program ill-formed.

Why does C++ require a user-provided default constructor to default-construct a const object?

The C++ standard (section 8.5) says:
If a program calls for the default initialization of an object of a const-qualified type T, T shall be a class type with a user-provided default constructor.
Why? I can't think of any reason why a user-provided constructor is required in this case.
struct B{
B():x(42){}
int doSomeStuff() const{return x;}
int x;
};
struct A{
A(){}//other than "because the standard says so", why is this line required?
B b;//not required for this example, just to illustrate
//how this situation isn't totally useless
};
int main(){
const A a;
}
The reason is that if the class doesn't have a user-defined constructor, then it can be POD, and the POD class is not initialized by default. So if you declare a const object of POD which is uninitialized, what use of it? So I think the Standard enforces this rule so that the object can actually be useful.
struct POD
{
int i;
};
POD p1; //uninitialized - but don't worry we can assign some value later on!
p1.i = 10; //assign some value later on!
POD p2 = POD(); //initialized
const POD p3 = POD(); //initialized
const POD p4; //uninitialized - error - as we cannot change it later on!
But if you make the class a non-POD:
struct nonPOD_A
{
nonPOD_A() {} //this makes non-POD
};
nonPOD_A a1; //initialized
const nonPOD_A a2; //initialized
Note the difference between POD and non-POD.
User-defined constructor is one way to make the class non-POD. There are several ways you can do that.
struct nonPOD_B
{
virtual void f() {} //virtual function make it non-POD
};
nonPOD_B b1; //initialized
const nonPOD_B b2; //initialized
Notice nonPOD_B doesn't defined user-defined constructor. Compile it. It will compile:
http://www.ideone.com/h7TsA
And comment the virtual function, then it gives error, as expected:
http://www.ideone.com/SWk7B
Well, I think, you misunderstood the passage. It first says this (§8.5/9):
If no initializer is specified for an object, and the object is of (possibly cv-qualified) non-POD class type (or array thereof), the object shall be default-initialized; [...]
It talks about non-POD class possibly cv-qualified type. That is, the non-POD object shall be default-initialized if there is no initializer specified. And what is default-initialized? For non-POD, the spec says (§8.5/5),
To default-initialize an object of type T means:
— if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);
It simply talks about default constructor of T, whether its user-defined or compiler-generated is irrelevant.
If you're clear up to this, then understand what the spec next says ((§8.5/9),
[...]; if the object is of const-qualified type, the underlying class type shall have a user-declared default constructor.
So this text implies, the program will be ill-formed if the object is of const-qualified POD type, and there is no initializer specified (because POD are not default initialized):
POD p1; //uninitialized - can be useful - hence allowed
const POD p2; //uninitialized - never useful - hence not allowed - error
By the way, this compiles fine, because its non-POD, and can be default-initialized.
Pure speculation on my part, but consider that other types have a similar restriction, too:
int main()
{
const int i; // invalid
}
So not only is this rule consistent, but it also (recursively) prevents unitialized const (sub)objects:
struct X {
int j;
};
struct A {
int i;
X x;
}
int main()
{
const A a; // a.i and a.x.j in unitialized states!
}
As for the other side of the question (allowing it for types with a default constructor), I think the idea is that a type with a user-provided default constructor is supposed to always be in some sensible state after construction. Note that the rules as they are allow for the following:
struct A {
explicit
A(int i): initialized(true), i(i) {} // valued constructor
A(): initialized(false) {}
bool initialized;
int i;
};
const A a; // class invariant set up for the object
// yet we didn't pay the cost of initializing a.i
Then perhaps we could formulate a rule like 'at least one member must be sensibly initialized in a user-provided default constructor', but that's way too much time spent trying to protect against Murphy. C++ tends to trust the programmer on certain points.
This was considered a defect (against all versions of the standard) and it was resolved by Core Working Group (CWG) Defect 253. The new wording for the standard states in http://eel.is/c++draft/dcl.init#7
A class type T is const-default-constructible if
default-initialization of T would invoke a user-provided constructor
of T (not inherited from a base class) or if
each direct non-variant non-static data member M of T has a default member initializer or, if M is of class type X (or array thereof), X
is const-default-constructible,
if T is a union with at least one non-static data member, exactly one variant member has a default member initializer,
if T is not a union, for each anonymous union member with at least one non-static data member (if any), exactly one non-static data
member has a default member initializer, and
each potentially constructed base class of T is const-default-constructible.
If a program calls for the default-initialization of an object of a
const-qualified type T, T shall be a const-default-constructible class
type or array thereof.
This wording essentially means that the obvious code works. If you initialize all of your bases and members, you can say A const a; regardless of how or if you spell any constructors.
struct A {
};
A const a;
gcc has accepted this since 4.6.4. clang has accepted this since 3.9.0. Visual Studio also accepts this (at least in 2017, not sure if sooner).
I was watching Timur Doumler's talk at Meeting C++ 2018 and I finally realised why the standard requires a user-provided constructor here, not merely a user-declared one. It has to do with the rules for value initialisation.
Consider two classes: A has a user-declared constructor, B has a user-provided constructor:
struct A {
int x;
A() = default;
};
struct B {
int x;
B() {}
};
At first glance, you might think these two constructors will behave the same. But see how value initialisation behaves differently, while only default initialisation behaves the same:
A a; is default initialisation: the member int x is uninitialised.
B b; is default initialisation: the member int x is uninitialised.
A a{}; is value initialisation: the member int x is zero-initialised.
B b{}; is value initialisation: the member int x is uninitialised.
Now see what happens when we add const:
const A a; is default initialisation: this is ill-formed due to the rule quoted in the question.
const B b; is default initialisation: the member int x is uninitialised.
const A a{}; is value initialisation: the member int x is zero-initialised.
const B b{}; is value initialisation: the member int x is uninitialised.
An uninitialised const scalar (e.g. the int x member) would be useless: writing to it is ill-formed (because it's const) and reading from it is UB (because it holds an indeterminate value). So this rule prevents you from creating such a thing, by forcing you to either add an initialiser or opt-in to the dangerous behaviour by adding a user-provided constructor.
I think it would be nice to have an attribute like [[uninitialized]] to tell the compiler when you're intentionally not initialising an object. Then we wouldn't be forced to make our class not trivially default constructible to get around this corner case. This attribute has actually been proposed, but just like all the other standard attributes, it does not mandate any normative behaviour, being merely a hint to the compiler.
Congratulations, you've invented a case in which there need not be any user defined constructor for the const declaration with no initializer to make sense.
Now can you come up with a reasonable re-wording of the rule that covers your case but still makes the cases that should be illegal illegal? Is it less than 5 or 6 paragraphs? Is it easy and obvious how it should be applied in any situation?
I posit that coming up with a rule that allows the declaration you created to make sense is really hard, and making sure that the rule can be applied in a way that makes sense to people when reading code is even harder. I would prefer a somewhat restrictive rule that was the right thing to do in most cases to a very nuanced and complex rule that was difficult to understand and apply.
The question is, is there a compelling reason the rule should be more complex? Is there some code that would otherwise be very difficult to write or understand that can be written much more simply if the rule is more complex?