I am learning about classes in C++ and know that non-static member functions have implicit this parameter. My first question is that does constructor also have an implicit this parameter just like non-static member functions. Note that i am not asking whether we can use this inside a ctor as i already know that we can use this inside a ctor.
Next, I know that inside a const qualified non-static member function for a class X, the type of this is const X*. And for a non-static member function(without const qualified), the type of this is X*. Similarly, inside the ctor the type of this is always X*. Here comes the deeper question.
We know that when we call a non-static member function(say like obj.func()), then the address of the object named obj is implicitly passed to the implicit this parameter of the method func. So this explains "where the this comes from in case of non-static member function".
Now, lets apply the same thing to constructors. For example, say we create an object of class X using default ctor like:
X x; //does the default ctor also have an implicit this parameter to which the address of x is passed?
My second question is: Does the same thing happen to ctors? Like, the address of x is passed to an implicit parameter of the default ctor. My current understanding is that ctors don't have an implicit this parameter. So when we write X x;, the address of x is not passed as an argument because the object is not created yet and so it doesn't make sense to pass its address. But from the standard we know that inside a ctor we can use the this pointer. So my second question essentially is that, if ctors don't have an implicit this parameter then where does the this in the statement this->p = 0; come from? We know that before using any name(like variable name) in C++, we must have a declaration for that name. So where is the declaration for this? Does the compiler implicitly declares the this in case of ctor? I mean, in case of non-static member function, i can understand that they have the declaration for this as the implicit this parameter but what happens in ctors? How inside ctors we're able to use the name this without having a declaration for that?
struct Name
{
private:
int p = 0;
int k = 0;
void func() //func is a non-static member function and so have an implicit this parameter
{
this->k = 0; // the "this" here comes from implicit this parameter
}
Name()
{
this->p = 0; //where does the "this" comes from here since ctor don't have implicit this parameter
}
};
My third question is that is the concept of implicit this parameter an implementation detail or does the standard says that non-static member function will have an implicit this parameter.
Summary
Do ctors have an implicit this parameter? This first question can also be phrased as "Do ctors also have an implicit object parameter?".
The standard says that we can use this inside a ctor. But where does that this come from. For example, in case of a non-static member function, we know that this comes from the implicit this parameter, but in case of ctors, since ctors don't have an implicit this parameter, where does that this that we're allowed to use inside ctor come from.
Is the concept of implicit this parameter an implementation detail or does the standard say that all non-static member functions have an implicit this parameter, in which case, ctors are also allowed by implementations to have an implicit this parameter.
Edit:
The most important part(IMO) of this question is that how are we able to use the name this inside a ctor? For example when we write:
this->p = 0; //here "this" behaves like a name
In the above statement this behaves like a name. And we know that before using any name(like variable name) in C++, we must have a declaration for that name. So where is the declaration for this? Does the compiler implicitly declares the this in case of ctor? I mean, in case of non-static member function, i can understand that they have the declaration for this as the implicit this parameter but what happens in ctors? How inside ctors we're able to use the name this without having a declaration for that?
From the perspective of the C++ standard
The standard only describes the semantics of the this keyword and not how its value gets there. That's completely abstracted away and an implementation has great flexibility in making it happen.
From the perspective of theoretical computer science
The address of the object under construction, available via the this keyword, absolutely is an input to the initialization procedure (called "constructor" in C++) for the object. So are addresses of virtual base subobjects, which C++ also makes available via the this keyword but cannot be calculated from the main input, so any such must be additional inputs.
Note that CS tends to use "parameter" with procedures more in the sense of template parameters than dynamic variable inputs. And CS uses "function" to mean a procedure without side effects. A C++ constructor is not a CS "function" and while templated constructors are possible (parametric procedures) the value of this is an ordinary input not a parameterization.
Neither is a C++ constructor a method -- there is no polymorphic target-type-dependent dispatch, and so in particular the this input is not used for dispatch.
From the perspective of ABI rules for parameter-passing
A C++ constructor is a special member function. There's no way to construct a function pointer for it or invoke it C-style; ABI requirements for C functions do not apply.
If the platform ABI explicitly describes C++ behaviors, then there will be one or more rules for passing the value(s) of this to a C++ constructor. Those rules may or may not specify a mechanism equivalent to other function arguments, but every ABI-compliant C++ compiler targeting that platform will pass this values as required by the special rules for constructors.
Notably, the ABI for passing this to a constructor isn't required to be equivalent to how this is passed to other (non-special and non-static) member functions, although practical ABIs may well use the same approach for both.
There's no such thing as "implicit this parameter" in the standard. The standard calls it an "implicit object parameter".
An implicit object parameter is only relevant to overload resolution, it doesn't "become" this. Separately, this is defined to have the same cv qualification as the member function.
Do ctors have an implicit this parameter? This first question can also be phrased as "Do ctors also have an implicit object parameter?".
No, from [over.match.funcs]
For the purposes of overload resolution, both static and non-static member functions have an implicit object parameter, but constructors do not.
But where does that this come from.
The object being constructed.
Is the concept of implicit this parameter an implementation detail or does the standard say that all non-static member functions have an implicit this parameter, in which case, ctors are also allowed by implementations to have an implicit this parameter.
The implicit object parameter is part of the rules of overload resolution, it doesn't influence this.
How inside ctors we're able to use the name this without having a declaration for that?
this is not a name, it is a language keyword. The language defines this in a non-static member function to be a prvalue expression. Unlike an unqualified-id (that names an object) in the same position, which are glvalue expressions.
That's the language-lawyer answer. Having said that, I do find that "implicit object parameter becomes this" is a useful mental model.
Recall that constructors (and destructors) can't be cv qualified, so there isn't anything for it to distinguish in a constructor, so it doesn't matter if it exists or not.
"this" is a keyword; it doesn't need declaring, but is always available in non-static member functions.
See e.g. the draft standard here.
The keyword this names a pointer to the object for which an implicit object member function is invoked or a non-static data member's initializer is evaluated.
Note that the mechanism behind this is unspecified.
Related
I recently learnt that constructors do not have names in C++ and some other things about them. I am also aware that a function has a type in C++ called a function type. For example,
void func(int)
{
}
In the above snippet the func has the function type void (int).
Now, I want to know that since constructors are special member functions then do they also have a type like the one shown above. For example say we have:
struct Name
{
Name(int)
{
}
};
Does the constructor shown above also has a function type just like ordinary functions or ordinary member functions. If yes, then how can we find that type. Like we can use decltype on ordinary functions, is it permitted to use decltype on constructors to find their type.
is it permitted to use decltype on constructors to find their type
It's not permitted. Primarily because there is no way to name a constructor. A common misnomer is that an expression like Name(0) or new Name(0), calls the constructor. But that isn't the case like in func(0). A constructor is never called by us directly, but rather always indirectly by the language construct that requires a new object to come into being.
[class.ctor.general]
1 ... Constructors do not have names.
2 A constructor is used to initialize objects of its class type. Because constructors do not have names, they are never found during name lookup; however an explicit type conversion using the functional notation ([expr.type.conv]) will cause a constructor to be called to initialize an object.
[Note 1: The syntax looks like an explicit call of the constructor. — end note]
Because we cannot name them, we cannot use introspection mechanisms like decltype to examine them. Therefore the standard doesn't specify a "type" for constructors, since there is no way for a strictly standard compliant program to examine said type.
A constructor also cannot possess a signature (as defined by the standard), since that by definition includes the function name (and constructors are, as mentioned, nameless).
[defns.signature.member] signature
⟨class member function⟩ name, parameter-type-list, class of which the function is a member, cv-qualifiers (if any), ref-qualifier (if any), and trailing requires-clause (if any)
A classes constructor is never called explicitly. You instantiate an object using something like new Name(5), memory will be allocated and then some part of that memory may be initialized by the steps defined in the constructors body.
Notice that a constructor has no return statement. What is returned by new Name(5) is a memory reference to the memory allocated by new.
This is given away by syntax like:
Name * foo = new Name(5)
foo is the pointer to whatever has be allocated and type checking can be done because Name referrs to a class, not to its constructor.
The specification does not seem to put any constraints on the member functions of a literal class type
I have two questions regarding this
Q1) Do I have complete liberty over what member functions I can put in?
Q2) How do I verify if a class is a literal type? (Possible method: Define a constexpr object of it and check if it compiles?)
The only constraints on literal classes I see are:
• All the data members must have literal type.
• The class must have at least one constexpr constructor.
• If a data member has an in-class initializer, the initializer for a member of built-in type must be a constant expression, or if the member has class type, the initializer must use the member’s own constexpr constructor.
• The class must use default definition for its destructor, which is the member that destroys objects of the class type
(Source: C++ Primer, 5th edition)
Q1. Yes, you can have any methods you like (excluding constructor/destructor which have constraints). Even including virtual methods, if the constructor is constexpr.
Q2. As you say, define a constexpr variable of that type. If there is no diagnostic message (and the compiler is conforming) then the type is definitely a LiteralType. Note that it is possible for the type to be literal but the code to fail compilation for some other reason.
The definition in the Standard seems slightly clearer to me than your quoted definition. For example, there are some cases where a constexpr constructor is not required (e.g. a closure or an aggregate).
I want to overload "->" operator globally for pointers. Is it possible?
A *var1 = new A(21);
int x = var1->someInt;
In that code it has to be triggered when reaching "var1"s "someInt" variable.
Actually I am trying to check the pointer before reaching its members. Also, I am trying to implement this system on a written code, so I don't want to change too much thing in the actual code.
I want to overload "->" operator globally. Is it possible?
No it is impossible.
Perhaps you might define a "root" class in your library or program (which would define that operator ->), and have all your classes inherit from it.
BTW, I am not sure you will be able to define your isPointerValid function in all cases (including the address of valid automatic variables of any class type), and efficiently.
There are many cases where you could not define such a function (e.g. union-s used in tagged union types; you don't easily know what member of a union is currently active. arbitrary casts; ...); .
For existing code and classes, the builtin meaning of -> (which you usually cannot redefine) has already been used to compile the code using them.
As described in C++17 standard draft in section 16.5.6 (emphasis mine):
An operator function shall either be a non-static member function or
be a non-member function that has at least one parameter whose type
is a class, a reference to a class, an enumeration, or a reference to
an enumeration.
Hence, it is not possible to overload an operator which doesn't take any arguments.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Nonstatic member as a default argument of a nonstatic member function
Correct me if I am wrong, but the way I think default parameters work is this:
When the compiler sees the function call, it starts pushing the parameters onto the stack. When it runs out of parameters, it will start pushing the defaults onto the stack until all required parameters are filled (I know this is a simplification, since parameters are actually pushed from right to left, so it will start with the defaults, but the idea is the same).
If this is true, why can't member variables be used as defaults? It seems to me that since the compiler is pushing them as usual at the call site, it should be able to resolve them just fine!
EDIT Since it seems by the answers my question was misunderstood, let me clarify. I know this is the case, and I know what is and isn't allowed by the language. My question is why did the language designers choose to not allow this, since it seems to naturally work.
The essence of what you are asking can be distilled into this simple example
void foo(int a, int b = a);
This is not allowed in C++. C++ does not allow default arguments to depend on other parameters.
Using class members as default arguments is just a particular case of the above, since class members are accessed through this pointer and this pointer is just another hidden parameter of each non-static member function.
So, the question is really why
void foo(int a, int b = a);
is not allowed.
One obvious potential reason to disallow this is that it would impose additional requirements on the order of argument evaluation. As you know, in C++ the order of function argument evaluation is unspecified - the compiler can evaluate arguments in any order. However, in order to support the above default argument functionality the compiler would have to make sure that a is evaluated before b. This feels like an excessive requirement, which restricts the typical freedom of evaluation order that we are used to seeing in C++.
Note that this
int a;
void foo(int b = a);
is allowed in C++. And, obviously, it does not exhibit the aforementioned order of evaluation issue.
I believe this are the most fitting paragraphs from the standard, especially §9:
8.3.6 Default arguments [dcl.fct.default]
§7 Local variables shall not be used in a default argument
§9 [...] Similarly, a non-static member shall not be used in a default argument, even if it is not evaluated, unless it appears as the id-expression of a class member access expression (5.2.5) or unless it is used to form a pointer to member (5.3.1).
This post lists all ways a default param can be set - Must default function parameters be constant in C++?
It's not difficult to workaround your need.
class A
{
int a;
public:
void f(int i);
void f()
{
f(a);
}
};
gives you what you want.
Summarizing Nawaz excellent answer in the linked question: The call to void Foo::Bar(int a = this->member) really means void Foo__Bar(Foo* this, int a = this->member). Obviously the second argument cannot be evaluated before the first, which violates a C++ axiom that compilers can evaluate arguments in whatever order they like.
What are the existing rules for taking function pointers or member function pointers to Standard functions? For example, something like
auto p = &std::string::size;
Is this legal? Would it be more or less legal if I explicitly requested the correct type, so it would function even if there was an additional implementation-added overload of std::string::size?
Using the "correct" type doesn't make things better: Except for the virtual functions all functions in the standard C++ library can have additional arguments as long as these are defaulted. Since the functions can also be declared with additional overloads (again with the exception of the virtual function), you can end up trying to assign an overload set a variable. Thus, the code isn't portable and there is no way to make it portable by using some sort of cast or some signature instead of auto.
The relevant quote is 17.6.5.5 [member.functions] paragraph 1:
An implementation may declare additional non-virtual member function signatures within a class:
--- by adding arguments with default values to a member function signature;
— by replacing a member function signature with default values by two or more member function signatures with equivalent behavior; and
— by adding a member function signature for a member function name.
I don't see a similar permission for non-member functions, though. Not sure where the permission to mess with these is hiding but I'm relatively sure that there are some weasel words for these as well. Looking further, it seems non-member functions are more constrained according to 17.6.5.4 [global.functions] paragraph 3:
An implementation shall not declare a global or non-member function signature with additional default arguments.
This would imply that you can take the address of the non-member functions, at least, when specifying the desired signature.