Related
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
name hiding and fragile base problem
I'm familiar with the rules involving member function hiding. Basically, a derived class with a function that has the same name as a base class function doesn't actually overload the base class function - it completely hides it.
struct Base
{
void foo(int x) const
{
}
};
struct Derived : public Base
{
void foo(const std::string& s) { }
};
int main()
{
Derived d;
d.foo("abc");
d.foo(123); // Will not compile! Base::foo is hidden!
}
So, you can get around this with a using declaration. But my question is, what is the reason for base class function hiding? Is this a "feature" or just a "mistake" by the standards committee? Is there some technical reason why the compiler can't look in the Base class for matching overloads when it doesn't find a match for d.foo(123)?
Name lookup works by looking in the current scope for matching names, if nothing is found then it looks in the enclosing scope, if nothing is found it looks in the enclosing scope, etc. until reaching the global namespace.
This isn't specific to classes, you get exactly the same name hiding here:
#include <iostream>
namespace outer
{
void foo(char c) { std::cout << "outer\n"; }
namespace inner
{
void foo(int i) { std::cout << "inner\n"; }
void bar() { foo('c'); }
}
}
int main()
{
outer::inner::bar();
}
Although outer::foo(char) is a better match for the call foo('c') name lookup stops after finding outer::inner::foo(int) (i.e. outer::foo(char) is hidden) and so the program prints inner.
If member function name weren't hidden that would mean name lookup in class scope behaved differently to non-class scope, which would be inconsistent and confusing, and make C++ even harder to learn.
So there's no technical reason the name lookup rules couldn't be changed, but they'd have to be changed for member functions and other types of name lookup, it would make compilers slower because they'd have to continue searching for names even after finding matching names in the current scope. Sensibly, if there's a name in the current scope it's probably the one you wanted. A call in a scope A probably wants to find names in that scope, e.g. if two functions are in the same namespace they're probably related (part of the same module or library) and so if one uses the name of the other it probably means to call the one in the same scope. If that's not what you want then use explicit qualification or a using declaration to tell the compiler the other name should be visible in that scope.
Is this a "feature" or just a "mistake" by the standards committee?
It's definitely not a mistake, since it's clearly stipulated in the standard. It's a feature.
Is there some technical reason why the compiler can't look in the Base class for matching overloads when it doesn't find a match for d.foo(123)?
Technically, a compiler could look in the base class. Technically. But if it did, it would break the rules set by the standard.
But my question is, what is the reason for base class function hiding?
Unless someone from the committee comes with an answer, I think we can only speculate. Basically, there were two options:
if I declare a function with the same name in a derived class, keep the base class's functions with the same name directly accessible through a derived class
don't
It could have been determined by flipping a coin (...ok, maybe not).
In general, what are the reasons for wanting a function with the same name as that of a base class? There's different functionality - where you'd more likely use polymorphism instead. For handling different cases (different parameters), and if these cases aren't present in the base class, a strategy pattern might be more appropriate to handle the job. So most likely function hiding comes in effect when you actually do want to hide the function. You're not happy with the base class implementation so you provide your own, with the option of using using, but only when you want to.
I think it's just a mechanism to make you think twice before having a function with the same name & different signature.
I believe #Lol4t0 is pretty much correct, but I'd state things much more strongly. If you allowed this, you'd end up with two possibilities: either make a lot of other changes throughout almost the entirety of the language, or else you end up with something almost completely broken.
The other changes you'd make to allow this to work would be to completely revamp how overloading is done -- you'd have to change at least the order of the steps that were taken, and probably the details of the steps themselves. Right now, the compiler looks up the name, then forms an overload set, resolves the overload, then checks access to the chosen overload.
To make this work even sort of well, you'd pretty much have to change that to check access first, and only add accessible functions to the overload set. With that, at least the example in #Lol4t0's answer could continue to compile, because Base::foo would never be added to the overload set.
That still means, however, that adding to the interface of the base class could cause serious problems. If Base didn't originally contain foo, and a public foo were added, then the call in main to d.foo() would suddenly do something entirely different, and (again) it would be entirely outside the control of whoever wrote Derived.
To cure that, you'd just about have to make a fairly fundamental change in the rules: prohibit implicit conversions of function arguments. Along with that, you'd change overload resolution so in case of a tie, the most derived/most local version of a function was favored over a less derived/outer scope. With those rules, the call to d.foo(5.0) could never resolve to Derived::foo(int) in the first place.
That, however, would only leave two possibilities: either calls to free functions would have different rules than calls to member functions (implicit conversions allowed only for free functions) or else all compatibility with C would be discarded entirely (i.e., also prohibit implicit conversions in all function arguments, which would break huge amounts of existing code).
To summarize: to change this without breaking the language entirely, you'd have to make quite a few other changes as well. It would almost certainly be possible to create a language that worked that way, but by the time you were done it wouldn't be C++ with one minor change -- it would be an entirely different language that wasn't much like C++ or C, or much of anything else.
I can only propose, that this decision was made to make things simpler.
Imagine, that derived function will overload base one. Then, does the following code should generate compilation error, or use Deriveds function?
struct Base
{
private:
void foo(float);
}
struct Derived: public Base
{
public:
void foo(int);
}
int main()
{
Derived d;
d.foo(5.0f);
}
According to existing behavior of overloads this should generate error.
Now imagine, in the first version Base had no foo(float). In second version it appears. Now changing the realization of base class breaks interface of derived.
If you are developer of Derived and cannot influence developers of Base and a lot of clients use your interface, you are in a bad situation now.
To my surprise, I learned that code like this actually works:
class A {
B b;
C c = this->b.GetC();
}
I guess I've been conditioned into thinking that this is essentially akin to self in Python. However, it also makes sense to me that at the class level we're declaring a template for which every instance of the class is to copy, and at the "declaration" level this shouldn't exist yet. But, C++ never fails to surprise, and the code above seems to work.
That said, is this bad style? What exactly are the differences between the code above, and simply initializing C in the constructor for A?
class A {
B b;
C c;
A() {
c = this->b.GetC();
}
}
(or simply A() : c(this->b.GetC()) {}).
What exactly are the differences between the code above, and simply initializing C in the constructor for A?
This is completely equivalent to A() : c(b.GetC()) {}. There is no difference.
The only difference you will observe is when using multiple constructors. It will use the constructor's initializer instead of the default one if specified.
the "declaration" level this shouldn't exist yet. But, C++ never fails to surprise, and the code above seems to work.
So this can exist at the class level in C++?
No. The initializer is ran when the constructor is called. The member to initialize only exist when an instance is created.
However, it also makes sense to me that at the class level we're declaring a template for which every instance of the class is to copy, and at the "declaration" level this shouldn't exist yet. But, C++ never fails to surprise, and the code above seems to work.
You are kind of right. If you had
struct foo
{
int bar;
decltype(this->bar) baz;
};
then you would get a compiler error for using this at the top level of the class.
In your case though using this like
C c = this->b.GetC();
//or
C c{this->b.GetC()};
isn't actually using this at the top level. A in class member initializer is just syntactic sugar for telling the class that if you do not manually initialize the member then you are to use the initializer provided. So you aren't really using this in the body of the class, you're just using a shortcut.
What exactly are the differences between the code above, and simply initializing C in the constructor for A?
The benefit is if you have some default value you want that memeber to have, you don't have to specify it in all of the constructors you create. You do it once, and your covered. If you ever need to change that value you can't mess it up because there is only one place to change it. You can't forget to do it in a particular constructor which is really nice.
Do note that there is a case where you have to use this in a in class member initializer. If the member you want to use comes from the dependent name, like a template base class, then you need to use this in order for the compiler to resolve the name.
I have a few questions/a general question regarding hiding?
A base class has a function: f() a derived class now declares virtual f()
What happens if the derived function is const? Does this not hide?
What about if the return types are different?
The parameters are different?
One is static and the other not?
Just wondering which of these differences between the two functions cause hiding/do not cause hiding.
Hiding is about names, not signatures.
First thing first.
Hiding refers to the idea that if you override a base class function in a derived class then in that derived class the original base class method that you defined will be hidden. -- That said, if you need to access your definition of the function from the base class then you'd need to have this in your code to it is NOT hidden :
using baseClass::functionName();
Const - if you define a function twice once with const another without const they are different. Scott Mayers book effective c++ contains an excellent chapter explaining const and non const-ness, I suggest you read it for a thorough understanding.
Your can not over-ride a function by having different return types, it will probably give compile time error stating its ambiguous - test it out, to make sure.
You CAN however overload a function by having different argument list, that is how one would overload a function.
If you're confused between over-ride and over-load - then I suggest you also read up on that. In a nutshell, overload is when you have two functions in the same scope with different argument lists, where as override is when you redefine a base class virtual function in a derived class with the same argument list.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
name hiding and fragile base problem
I'm familiar with the rules involving member function hiding. Basically, a derived class with a function that has the same name as a base class function doesn't actually overload the base class function - it completely hides it.
struct Base
{
void foo(int x) const
{
}
};
struct Derived : public Base
{
void foo(const std::string& s) { }
};
int main()
{
Derived d;
d.foo("abc");
d.foo(123); // Will not compile! Base::foo is hidden!
}
So, you can get around this with a using declaration. But my question is, what is the reason for base class function hiding? Is this a "feature" or just a "mistake" by the standards committee? Is there some technical reason why the compiler can't look in the Base class for matching overloads when it doesn't find a match for d.foo(123)?
Name lookup works by looking in the current scope for matching names, if nothing is found then it looks in the enclosing scope, if nothing is found it looks in the enclosing scope, etc. until reaching the global namespace.
This isn't specific to classes, you get exactly the same name hiding here:
#include <iostream>
namespace outer
{
void foo(char c) { std::cout << "outer\n"; }
namespace inner
{
void foo(int i) { std::cout << "inner\n"; }
void bar() { foo('c'); }
}
}
int main()
{
outer::inner::bar();
}
Although outer::foo(char) is a better match for the call foo('c') name lookup stops after finding outer::inner::foo(int) (i.e. outer::foo(char) is hidden) and so the program prints inner.
If member function name weren't hidden that would mean name lookup in class scope behaved differently to non-class scope, which would be inconsistent and confusing, and make C++ even harder to learn.
So there's no technical reason the name lookup rules couldn't be changed, but they'd have to be changed for member functions and other types of name lookup, it would make compilers slower because they'd have to continue searching for names even after finding matching names in the current scope. Sensibly, if there's a name in the current scope it's probably the one you wanted. A call in a scope A probably wants to find names in that scope, e.g. if two functions are in the same namespace they're probably related (part of the same module or library) and so if one uses the name of the other it probably means to call the one in the same scope. If that's not what you want then use explicit qualification or a using declaration to tell the compiler the other name should be visible in that scope.
Is this a "feature" or just a "mistake" by the standards committee?
It's definitely not a mistake, since it's clearly stipulated in the standard. It's a feature.
Is there some technical reason why the compiler can't look in the Base class for matching overloads when it doesn't find a match for d.foo(123)?
Technically, a compiler could look in the base class. Technically. But if it did, it would break the rules set by the standard.
But my question is, what is the reason for base class function hiding?
Unless someone from the committee comes with an answer, I think we can only speculate. Basically, there were two options:
if I declare a function with the same name in a derived class, keep the base class's functions with the same name directly accessible through a derived class
don't
It could have been determined by flipping a coin (...ok, maybe not).
In general, what are the reasons for wanting a function with the same name as that of a base class? There's different functionality - where you'd more likely use polymorphism instead. For handling different cases (different parameters), and if these cases aren't present in the base class, a strategy pattern might be more appropriate to handle the job. So most likely function hiding comes in effect when you actually do want to hide the function. You're not happy with the base class implementation so you provide your own, with the option of using using, but only when you want to.
I think it's just a mechanism to make you think twice before having a function with the same name & different signature.
I believe #Lol4t0 is pretty much correct, but I'd state things much more strongly. If you allowed this, you'd end up with two possibilities: either make a lot of other changes throughout almost the entirety of the language, or else you end up with something almost completely broken.
The other changes you'd make to allow this to work would be to completely revamp how overloading is done -- you'd have to change at least the order of the steps that were taken, and probably the details of the steps themselves. Right now, the compiler looks up the name, then forms an overload set, resolves the overload, then checks access to the chosen overload.
To make this work even sort of well, you'd pretty much have to change that to check access first, and only add accessible functions to the overload set. With that, at least the example in #Lol4t0's answer could continue to compile, because Base::foo would never be added to the overload set.
That still means, however, that adding to the interface of the base class could cause serious problems. If Base didn't originally contain foo, and a public foo were added, then the call in main to d.foo() would suddenly do something entirely different, and (again) it would be entirely outside the control of whoever wrote Derived.
To cure that, you'd just about have to make a fairly fundamental change in the rules: prohibit implicit conversions of function arguments. Along with that, you'd change overload resolution so in case of a tie, the most derived/most local version of a function was favored over a less derived/outer scope. With those rules, the call to d.foo(5.0) could never resolve to Derived::foo(int) in the first place.
That, however, would only leave two possibilities: either calls to free functions would have different rules than calls to member functions (implicit conversions allowed only for free functions) or else all compatibility with C would be discarded entirely (i.e., also prohibit implicit conversions in all function arguments, which would break huge amounts of existing code).
To summarize: to change this without breaking the language entirely, you'd have to make quite a few other changes as well. It would almost certainly be possible to create a language that worked that way, but by the time you were done it wouldn't be C++ with one minor change -- it would be an entirely different language that wasn't much like C++ or C, or much of anything else.
I can only propose, that this decision was made to make things simpler.
Imagine, that derived function will overload base one. Then, does the following code should generate compilation error, or use Deriveds function?
struct Base
{
private:
void foo(float);
}
struct Derived: public Base
{
public:
void foo(int);
}
int main()
{
Derived d;
d.foo(5.0f);
}
According to existing behavior of overloads this should generate error.
Now imagine, in the first version Base had no foo(float). In second version it appears. Now changing the realization of base class breaks interface of derived.
If you are developer of Derived and cannot influence developers of Base and a lot of clients use your interface, you are in a bad situation now.
What is the difference between member and non-member functions in C++?
There are several differences between a member function (which I will now call method) and a free function (which I will now call function).
First, let's just state that they are not so different. Object code can generally be compiled down to C (or assembly) which are procedural languages with no notion of methods. Both methods and functions are then called like subroutines.
Now that this is out of the way, let's look at the differences. They can be classified in two categories: conceptual and syntactic.
Syntactically
The syntax is the obvious part of any language, so it's the easiest to get out of the way.
First note: there are two different kinds of methods in C++ (and a number of other languages), the static methods and regular methods.
Both kinds of methods have full access to the class internals (protected and private sections) as well (of course) as access to the class public interface.
static methods are equivalent to friend functions (apart from some scoping differences).
Within a regular method, a special keyword (this in C++) allows access to the current object on which the method has been invoked (via the ., ->, .* or ->* operators).
A regular method may be const and/or volatile qualified, enabling it for (respectively) const and/or volatile qualified objects. For example, a non-const method cannot be called on a const object. This can be seen as qualifying this within the method, ie void Foo::bar() const has a this of type Foo const*.
A regular method may be marked as virtual. Virtuality enables runtime polymorphism by enabling overriding. I won't extend on this mechanism here, let's just note that functions cannot be virtual.
One often ignored point, is that methods (both static and regular) are scoped within the class. This is important for name lookup (of other methods or attributes/variables) as it means that the elements of the class have priority during lookup from a method on the elements declared outside of the class.
Since the qualification using this-> before attribute or methods is not mandatory, this comes handy in regular methods, though it may introduce subtle bugs. In static methods, it avoids qualifying by the class name the static attributes and methods one whishes to access.
Now that the main syntactic differences have been asserted, let's check the conceptual differences.
Conceptually
OOP is generally about tying together state and behavior (of this state). This is done by creating classes which group attributes (state) and behavior (methods) and (in theory) stating that only the methods can act on the state. Therefore, in OOP, the methods are responsible for implementing the behavior of the class.
The methods participate to the encapsulation of state (freeing clients from the implementation detail) and to the preservation of the class invariants (statements about the class state that hold true from its birth to its death, whatever you do with it).
C++
In C++, as we have seen previously, this is done by using different access levels (public, protected and private) and granting access to the non-public levels to a restricted portion of the code. Generally attributes will be private and thus only accessible to the class methods (and maybe some friends, for syntax quirks).
Note: I urge you not to use protected attributes, it's hard to track down their modifications and since the set of derived classes is unbounded... their implementation cannot be changed easily afterward.
However, beware that C++ discourages from bloating the interface with lots of methods.
The trouble is that because methods are responsible of maintaining invariants, the more there are and the more the responsability is spread, making it more difficult to track down bugs and ensure correctness. Also, since methods depends on the class internals, it makes change more costly.
Instead, in C++, it is generally advised to write a minimal set of methods and delegating the rest of the behavior to non-friend functions (as long as it doesn't increase the cost too much).
See Sutter's take on std::string in Monolith Unstrung.
The delegation to non-friend methods was emphasized by Sutter in its Interface Principle in which he states that functions that are delivered with the class (in the same file/same namespace) and use the class, are logically part of the class interface. He restates in Exceptional C++.
This answer is becoming rather long-winded, yet I suspect that I have overlooked differences that other would find critical... oh well.
A (non-static) member function has an implicit this argument, a non-member doesn't.
Syntactically, you pass that implicit argument on the left of the . or -> operator like.so() or like->so(), instead of as a function argument so( like ).
Likewise, when declaring a member function, you need to do so in the class of which it is a member:
class Class {
public:
void a_public_member_function();
};
Non-member functions are instead declared outside any class (C++ calls this "at namespace scope").
(Non-static) member functions can also be virtual, but non-member functions (and static member functions) cannot.
A non-static member function is invoked on objects of the class it belongs to. It has implicitly access to the this pointer representing the current object. Through this pointer, it may access other members easily and with full access privileges (i.e. access private members).
A non-member function has no implicit this. In the sample below, bar is a member function while freebar is not. Both do more or less the same, but note how bar gets an implicit object pointer via this (also only bar has privileged access to foo's members, freebar only has access to public members).
class foo {
public:
void bar() {
this->x = 0; // equivalent to x = 0;
}
int x;
};
void freebar(foo* thefoo) {
thefoo->x = 1;
}
// ...
foo f;
f.bar();
// f.x is now 0
freebar(&f);
// f.x is now 1
Semantically a member function is more than just a function with an implicit this parameter. It is meant to define the behaviour of an object (i.e. a car object would have drive(), stop() as member functions).
Note that there are also static member functions which have full privileges but don't get an implicit this nor are they invoked through an instance of the class (but rather through the full name of the class).
In the following code, f() is a member function of class Sample, and g() is a non-member function:
class Sample
{
void f();
};
void g();
Its very simple. Since f() is a member of the class Sample, so its called member function (of class Sample). And since g() is not member of any class, so its called non-member function.
A member function is invoked on an object and has access to the fields of the class.
Member functions can be polymorphic (via the virtual keyword) which is essential for OOP.
Member functions get called on instances and have a this pointer available; non-members don't.