The compiler inserts code into the beginning of the constructor that initializes the VPTR. Is this operation influenced by whether the constructor is the default one or it is explicitly declared? Is there any difference between the two?
It will always be inserted, it has to be for polymorphism to work, at least the way compilers typically have it implemented. It will be initialized always, the constructors you declare (or don't declare) are irrelevant.
Or, hey, maybe it won't (although I do not know any compilers that behave this way), all you need to care about is that polymorphism works, and will always work no matter what constructors you have declared.
Related
As far as I know, this initialization will be complemented before the function body of the constructor.
If the initialization of data members is the part of constructor, then it should be inlined when the contructor is inlined, otherwise on the contrary.
By the way, how about the constructor list? And is the destroy of data member part of deconstructor?
The assumption is wrong. Compilers may partially inline functions. It's even easier to do so for constructors and destructors as you can't take their address anyway.
Still, that's something you won't notice at C++ level, only if you inspect the assembly.
You can consider the member initialization part of the constructor for functional effects and the member destruction part of the destructor. But what is inlined and what is invoked is not guaranteed. This part is depends totally on the compiler you are using.
Some days ago, while reading Standard C++ news I've read the post about Defaulted functions in C++11, in that article is mentioned that the user-defined constructor is less efficient than the one generated by the compiler:
The user-defined default constructor is less efficient than the compiler implicitly defined default constructor.
Continuing the reading, there's an example where an user-defined constructor is marked as default, and then says:
the explicitly defaulted constructor is more efficient than a manually programmed default constructor.
I don't understand these assertions, so I was wondering:
Why a user-default constructor (or special member function) would be less efficient than the compiler implicitly defined one?
How is the efficiency improved by explicitly defaulting a constructor (or special member function)?
What guidelines I must follow to choose to default a constructor (or special member function) and how the efficiency affects this decision?
I think a better statement is that a user-defined default constructor MAY be less efficient than a compiler generated out.
For example, when it's generating a default constructor internally the compiler may be able to make assumptions and optimizations that it can't make for a user-defined contstructor (side-effects come to mind).
Also keep in mind that a user-defined default constructor could do totally different work that default-constructing all its members, resulting in it being less efficient (but also more correct). This doesn't seem to be the case in the link you provided however.
And we all know, that if it's written on internet then it is must be right... Wait, do we?
In the article where I found the first assertion of less efficient, the author tells the truth. Though you seem to misinterpret it -- in the example it refers to the hand-crafted ctor uses assignment. For no good reasons, and going against 2 decade old guidelines.
Next instance, same case. (As a practical note I shall add, that for any compiler claiming to have optimizations I expect the same assy output even for that form...)
I see no reason why the proper handwritten ctor would be different from the defaulted one in any ways, including efficiency. OTOH if they are identical why on earth write it? I'm all too happy that compiler makes it for me. And finally I can even control it in some ways I previously could not. Could use more of such functions. ;-)
Some of my colleagues prefer to explicitly initialize std::auto_ptr to 0 in constructor initialization list, but it will be initialized to 0 in it's constructor without any explicit initialization. So is there any reason to do it?
#include <memory>
class A
{
A() : SomePtr(0)
{
}
private:
std::auto_ptr<SomeType> SomePtr;
};
No, the default constructor of std::auto_ptr does exactly that, so doing it explicitly is not necessary. In any case, it's a matter of style and you should be consistent. For instance, would you explicitly call the default constructor of a member vector in the constructor initialization list?
As a side note, std::auto_ptr is deprecated in the upcoming standard
Psychology.
For built-in types, you probably already know they are uninitialized unless you do so explicitly. For classes, this is not the case.
A strive to consistency results in explicit initialization everywhere. This allows you to forget if A::SomePtr is a built-in or a class type. Pretty useless, imho, since the amount of built-in types is quite limited.
One reason maybe clarity, but that should be the only one. I myself prefer not to write unneccessary intialization, especially if that completely spares me from writing a default constructor for the surrounding class and just let the compiler do its job. Whereas it's merely a matter of style, I think too much over-paranoia does even harm the clarity of the code.
Could someone please explain what is meant by the following?
You must define a default constructor if your class defines member variables and has no other constructors. Otherwise the compiler will do it for you, badly.
What are they referring to as "badly"?
From the expansion of that link:
"The reason for this is that if you
have no other constructors and do not
define a default constructor, the
compiler will generate one for you.
This compiler generated constructor
may not initialize your object
sensibly."
Might refer to how new T and new T() differ when there is no ctor provided.
It's good to be sure that the object is created in a known state. Primitive variables won't be set to zero by default, so you could end up with subtle bugs that don't always show up. By initializing the member variables to sensible variables, everything is much more predictable.
The only problem with the default constructor is that it initializes only what the compiler thinks must be initialized, and not what you may think needs to be initialized. Basically, that means that it will invoke initializers for objects with default initializers. It won't set pointers or simple types like int to sane values, etc. If that is sufficient, then the default constructor is not 'bad'. When it is insufficient, it is a bug (in your code) that you did not define the necessary default constructor with the correct initialization.
Take the Google style guide with a grain of salt -- or maybe a truckload of salt.
It is true that the compiler-generated default constructor won't necessarily initialize members that are of built-in types in a meaningful fashion. If you want that done, then yes, its failure to do that is bad. OTOH, if you don't want that done, then its doing it could be somewhat bad (wasteful) as well.
Bottom line: there are times to write your own default ctor, but they tend toward the exception, not the rule. Although there are simple rules of thumb to cover a lot of cases in C++ and will prevent a lot of problems, this really isn't one of them -- here you pretty much do need to know what the compiler-generated ctor will do, and what you want different if you're going to write your own.
In Debug build most compilers fill uninitialized space with some magic values, so that debugging is reliable. And providing custom constructor prevents certain POD optimizations.
In fact, it's a guideline just to make sure people does explicitely make the statement of what is an invalid or default state of any object.
That way, no surprise when reading the code, compared to the actual execution.
However, think that it's a company guideline, that is used to make sure everyone does follow the same rules, that's not a you-must-follow-it-because-google-does-it.
In fact, if you manage to make all your member objects being in valid state when default constructed, or force you to set a constructor, then there is no good reason for such a guideline.
If you have any primitive types as member variables (eg. int, float), then the default ctor will not initialize them. member variables that are a class type will have their default ctor's invoked.
Prefer member initializer lists, so your user supplied ctor may be empty:
class Foo {
int bar;
float baz;
Foo(): bar(0), baz(0.0f) { /* empty ctor body */ }
};
It won't set integers to 0 or pointers to null. It will run default constructors on object of types with constructors.
Some people would call it 'not sensible'.
It just seems a too simplified version of the rules of 3, you should either define yourself or leave the compiler version of
the copy constructor
the assignment operator
the destructor
(Note that by defining yourself a copy constructor, the compiler won't define a default constructor).
The default constructor built by the compiler does 'nothing', it will not even zero the memory occupied by the object
As I understand the standard, a trivial destructor is one which is implicitly declared and whose class has only base and non-static members with trivial destructors.
Given the recursivity of this definition, it seems to me that the only "recursion-stopping" condition is to find a base or non-static member with a non-implicitly declared destructor (i.e. user declared).
If that's right, that should mean that a trivial destructor is one which "doesn't have to do anything" and hence it will be declared (implicitly) but not defined.
Saying it in another way: is it correct to say that an implicitly defined destructor (that is "it does something") cannot be trivial as per the standard definition?
Sorry for the kind of silly question, but I'd like to clarify things a bit in my head...
No. An implicitly defined, trivial destructor is by definition trivial :) The difference between the declare and define thingy is that in order for the compiler to even see that a destructor is available, there must always a declaration. So if you don't provide one, it will implicitly provide one.
But now, it will also define one, if that is needed (if an object of that class type is destroyed). In any case, it has to do something: It needs to call the destructors of all its members and base classes. A simple example which illustrates the effect of implicitly defining a destructor:
struct a {
private:
~a();
};
struct bug {
// note: can't be destructed
a a_;
};
As soon as you try to create a local object of bug, the compiler will signal an error, because it yields a definition of a destructor for bug, which tries to call the not accessible destructor of a.
Now, i think triviality of destructors/constructors are mostly used to put constraints on your program. Objects having non-trivial versions of them can't be put in unions, for example. On the other side, you can delete an object having incomplete type, provided it has a trivial destructor. Note that if your program can't decide whether or not a trivial destructor was actually defined, the compiler is allowed to omit defining it. That's the so-called as-if rule. The compiler has to behave as-if it's Standard compliant - optimizations do not matter as long as they don't change the meaning of a program.
Your wording is a bit unfortunate. E.g. the recursion of course also ends when you run out of members and base classes. Those wording problems also seem to get you more confused.
Anyway, all implicitly-declared destructors, whether they are trivial or not, are defined if and only if they are used. Used is a specific term here. A destructor of type T is used whenever the lifetime of a T object ends.
Trivial destructors exist because C programmers put structs in unions. This code should remian legal in C++, so the notion of a trivial destructor was invented for C++. All C structs have trivial destructors, when compiled as C++.
Consider these two classes:
class A {
};
class B {
private:
A obj;
};
The destructors of both these classes implicitly defined. Yet, at the same time, both of them are trivial by the standard definition.