I have a class with a method (getter) that performs some relatively expensive operation, so I want to cache its result. Calling the method does not change behavior of the object, but it needs to store its result in this, so it can't be const.
The problem is that I now have another method that is const and I need to invoke my getter. Is there some generally accepted solution to this problem? Should I bypass the constness checking in the getter to make it const, (Coudn't that cause problems with optimizing compiler?) or I have to propagate the non-constness to all methods that use this getter?
Example:
class MyClass
{
public:
Foo &expensiveGetter() /*const?*/
{
if (cachedValue == nullptr) {
cachedValue = computeTheValue();
}
return *cachedValue;
}
void myMethod() /* How to make this const? */
{
auto &foo = expensiveGetter();
// etc.
}
private:
Foo *cachedValue = nullptr;
}
I am looking for something like the RefCell in Rust.
This is one of the situations the mutable specifier fits particularly well. When a class member is mutable, that member can be changed even if the enclosing class is const.
So, if MyClass::cachedValue is a mutable Foo* instead of a Foo*, you can have const member functions in MyClass that make changes to cachedValue and then all calling code can just act on a const MyClass as normal.
Related
I'm working on learning C++ with Stroustrup's (Programming Principles & Practice Using C++) book. In an exercise we define a simple struct:
template<typename T>
struct S {
explicit S(T v):val{v} { };
T& get();
const T& get() const;
void set(T v);
void read_val(T& v);
T& operator=(const T& t); // deep copy assignment
private:
T val;
};
We're then asked to define a const and a non-const member function to get val.
I was wondering: Is there any case where it makes sense to have non-const get function that returns val?
It seems much cleaner to me that we can't change the value in such situations indirectly. What might be use cases where you need a const and a non-const get function to return a member variable?
Non-const getters?
Getters and setters are merely convention. Instead of providing a getter and a setter, a sometimes used idiom is to provide something along the line of
struct foo {
int val() const { return val_; }
int& val() { return val_; }
private:
int val_;
};
Such that, depending on the constness of the instance you get a reference or a copy:
void bar(const foo& a, foo& b) {
auto x = a.val(); // calls the const method returning an int
b.val() = x; // calls the non-const method returning an int&
};
Whether this is good style in general is a matter of opinion. There are cases where it causes confusion and other cases where this behaviour is just what you would expect (see below).
In any case, it is more important to design the interface of a class according to what the class is supposed to do and how you want to use it rather than blindly following conventions about setters and getters (eg you should give the method a meaningful name that expresses what it does, not just in terms of "pretend to be encapsulated and now provide me access to all your internals via getters", which is what using getters everywhere actually means).
Concrete example
Consider that element access in containers is usually implemented like this. As a toy example:
struct my_array {
int operator[](unsigned i) const { return data[i]; }
int& operator[](unsigned i) { return data[i]; }
private:
int data[10];
};
It is not the containers job to hide the elements from the user (even data could be public). You dont want different methods to access elements depending on whether you want to read or write the element, hence providing a const and a non-const overload makes perfectly sense in this case.
non-const reference from get vs encapsulation
Maybe not that obvious, but it is a bit controversial whether providing getters and setters supports encapsulation or the opposite. While in general this matter is to a large extend opinion based, for getters that return non const references it is not so much about opinions. They do break encapuslation. Consider
struct broken {
void set(int x) {
counter++;
val = x;
}
int& get() { return x; }
int get() const { return x; }
private:
int counter = 0;
int value = 0;
};
This class is broken as the name suggests. Clients can simply grab a reference and the class has no chance to count the number of times the value is modified (as the set suggests). Once you return a non-const reference then regarding encapsulation there is little difference to making the member public. Hence, this is used only for cases where such behaviour is natural (eg container).
PS
Note that your example returns a const T& rather than a value. This is reasonable for template code, where you dont know how expensive a copy is, while for an int you wont gain much by returning a const int& instead of an int. For the sake of clarity I used non-template examples, though for templated code you would probably rather return a const T&.
First let me rephrase your question:
Why have a non-const getter for a member, rather than just making the member public?
Several possible reasons reasons:
1. Easy to instrument
Whoever said the non-const getter needs to be just:
T& get() { return val; }
? it could well be something like:
T& get() {
if (check_for_something_bad()) {
throw std::runtime_error{
"Attempt to mutate val when bad things have happened");
}
return val;
}
However, as #BenVoigt suggests, it is more appropriate to wait until the caller actually tries to mutate the value through the reference before spewing an error.
2. Cultural convention / "the boss said so"
Some organizations enforce coding standards. These coding standards are sometimes authored by people who are possibly overly-defensive. So, you might see something like:
Unless your class is a "plain old data" type, no data members may be public. You may use getter methods for such non-public members as necessary.
and then, even if it makes sense for a specific class to just allow non-const access, it won't happen.
3. Maybe val just isn't there?
You've given an example in which val actually exists in an instance of the class. But actually - it doesn't have to! The get() method could return some sort of a proxy object, which, upon assignment, mutation etc. performs some computation (e.g. storing or retrieving data in a database; or flipping a bit, which itself is not addressable like an object needs to be).
4. Allows changing class internals later without changing user code
Now, reading items 1. or 3, above, you might ask "but my struct S does have val!" or "by my get() doesn't do anything interesting!" - well, true, they don't; but you might want to change this behavior in the future. Without a get(), all of your class' users will need to change their code. With a get(), you only need to make changes to the implementation of struct S.
Now, I don't advocate for this kind of a design approach approach, but some programmers do.
get() is callable by non const objects which are allowed to mutate, you can do:
S r(0);
r.get() = 1;
but if you make r const as const S r(0), the line r.get() = 1 no longer compile, not even to retrieve the value, that's why you need a const version const T& get() const to at least to able to retrieve the value for const objects, doing so allows you do:
const S r(0)
int val = r.get()
The const version of member functions try to be consistent with the constness property of the object the call is made on, i.e if the object is immutable by being const and the member function returns a reference, it may reflect the constness of the caller by returning a const reference, thus preserving the immutability property of the object.
It depends on the purpose of S. If it's some kind of a thin wrapper, it might be appropriate to allow the user to access the underlaying value directly.
One of the real-life examples is std::reference_wrapper.
No. If a getter simply returns a non-const reference to a member, like this:
private:
Object m_member;
public:
Object &getMember() {
return m_member;
}
Then m_member should be public instead, and the accessor is not needed. There is absolutely no point making this member private, and then create an accessor, which gives all access to it.
If you call getMember(), you can store the resulting reference to a pointer/reference, and afterwards, you can do whatever you want with m_member, the enclosing class will know nothing about it. It's the same, as if m_member had been public.
Note, that if getMember() does some additional task (for example, it doesn't just simply return m_member, but lazily constructs it), then getMember() could be useful:
Object &getMember() {
if (!m_member) m_member = new Object;
return *m_member;
}
I'm working on learning C++ with Stroustrup's (Programming Principles & Practice Using C++) book. In an exercise we define a simple struct:
template<typename T>
struct S {
explicit S(T v):val{v} { };
T& get();
const T& get() const;
void set(T v);
void read_val(T& v);
T& operator=(const T& t); // deep copy assignment
private:
T val;
};
We're then asked to define a const and a non-const member function to get val.
I was wondering: Is there any case where it makes sense to have non-const get function that returns val?
It seems much cleaner to me that we can't change the value in such situations indirectly. What might be use cases where you need a const and a non-const get function to return a member variable?
Non-const getters?
Getters and setters are merely convention. Instead of providing a getter and a setter, a sometimes used idiom is to provide something along the line of
struct foo {
int val() const { return val_; }
int& val() { return val_; }
private:
int val_;
};
Such that, depending on the constness of the instance you get a reference or a copy:
void bar(const foo& a, foo& b) {
auto x = a.val(); // calls the const method returning an int
b.val() = x; // calls the non-const method returning an int&
};
Whether this is good style in general is a matter of opinion. There are cases where it causes confusion and other cases where this behaviour is just what you would expect (see below).
In any case, it is more important to design the interface of a class according to what the class is supposed to do and how you want to use it rather than blindly following conventions about setters and getters (eg you should give the method a meaningful name that expresses what it does, not just in terms of "pretend to be encapsulated and now provide me access to all your internals via getters", which is what using getters everywhere actually means).
Concrete example
Consider that element access in containers is usually implemented like this. As a toy example:
struct my_array {
int operator[](unsigned i) const { return data[i]; }
int& operator[](unsigned i) { return data[i]; }
private:
int data[10];
};
It is not the containers job to hide the elements from the user (even data could be public). You dont want different methods to access elements depending on whether you want to read or write the element, hence providing a const and a non-const overload makes perfectly sense in this case.
non-const reference from get vs encapsulation
Maybe not that obvious, but it is a bit controversial whether providing getters and setters supports encapsulation or the opposite. While in general this matter is to a large extend opinion based, for getters that return non const references it is not so much about opinions. They do break encapuslation. Consider
struct broken {
void set(int x) {
counter++;
val = x;
}
int& get() { return x; }
int get() const { return x; }
private:
int counter = 0;
int value = 0;
};
This class is broken as the name suggests. Clients can simply grab a reference and the class has no chance to count the number of times the value is modified (as the set suggests). Once you return a non-const reference then regarding encapsulation there is little difference to making the member public. Hence, this is used only for cases where such behaviour is natural (eg container).
PS
Note that your example returns a const T& rather than a value. This is reasonable for template code, where you dont know how expensive a copy is, while for an int you wont gain much by returning a const int& instead of an int. For the sake of clarity I used non-template examples, though for templated code you would probably rather return a const T&.
First let me rephrase your question:
Why have a non-const getter for a member, rather than just making the member public?
Several possible reasons reasons:
1. Easy to instrument
Whoever said the non-const getter needs to be just:
T& get() { return val; }
? it could well be something like:
T& get() {
if (check_for_something_bad()) {
throw std::runtime_error{
"Attempt to mutate val when bad things have happened");
}
return val;
}
However, as #BenVoigt suggests, it is more appropriate to wait until the caller actually tries to mutate the value through the reference before spewing an error.
2. Cultural convention / "the boss said so"
Some organizations enforce coding standards. These coding standards are sometimes authored by people who are possibly overly-defensive. So, you might see something like:
Unless your class is a "plain old data" type, no data members may be public. You may use getter methods for such non-public members as necessary.
and then, even if it makes sense for a specific class to just allow non-const access, it won't happen.
3. Maybe val just isn't there?
You've given an example in which val actually exists in an instance of the class. But actually - it doesn't have to! The get() method could return some sort of a proxy object, which, upon assignment, mutation etc. performs some computation (e.g. storing or retrieving data in a database; or flipping a bit, which itself is not addressable like an object needs to be).
4. Allows changing class internals later without changing user code
Now, reading items 1. or 3, above, you might ask "but my struct S does have val!" or "by my get() doesn't do anything interesting!" - well, true, they don't; but you might want to change this behavior in the future. Without a get(), all of your class' users will need to change their code. With a get(), you only need to make changes to the implementation of struct S.
Now, I don't advocate for this kind of a design approach approach, but some programmers do.
get() is callable by non const objects which are allowed to mutate, you can do:
S r(0);
r.get() = 1;
but if you make r const as const S r(0), the line r.get() = 1 no longer compile, not even to retrieve the value, that's why you need a const version const T& get() const to at least to able to retrieve the value for const objects, doing so allows you do:
const S r(0)
int val = r.get()
The const version of member functions try to be consistent with the constness property of the object the call is made on, i.e if the object is immutable by being const and the member function returns a reference, it may reflect the constness of the caller by returning a const reference, thus preserving the immutability property of the object.
It depends on the purpose of S. If it's some kind of a thin wrapper, it might be appropriate to allow the user to access the underlaying value directly.
One of the real-life examples is std::reference_wrapper.
No. If a getter simply returns a non-const reference to a member, like this:
private:
Object m_member;
public:
Object &getMember() {
return m_member;
}
Then m_member should be public instead, and the accessor is not needed. There is absolutely no point making this member private, and then create an accessor, which gives all access to it.
If you call getMember(), you can store the resulting reference to a pointer/reference, and afterwards, you can do whatever you want with m_member, the enclosing class will know nothing about it. It's the same, as if m_member had been public.
Note, that if getMember() does some additional task (for example, it doesn't just simply return m_member, but lazily constructs it), then getMember() could be useful:
Object &getMember() {
if (!m_member) m_member = new Object;
return *m_member;
}
Having the following object
class Parser
{
public:
Parser(ComponentFactory * const factory): _factory(factory) {};
~Parser() = default;
void parse() const {
_factory->setFoo("foo");
}
private:
Factory * _factory;
};
My function parse() is specified as const. That's to say the function shouldn't modify the current object state and performs only read-only logics.
However, does the modification of the factory object imply a change of my current object's state? In other therms, is this even compilable?
I would like to understand why if yes, since I can't find any related subject on the net..
EDIT:
Since none can understand this, let me try to explain it better. In simple words, is the above code supposed to compile?
However, does the modification of the factory object imply a change of my current object's state?
No
In other therms, is this even compilable?
Yes.
I would like to understand why if yes, since I can't find any related subject on the net..
Say you have
struct Foo
{
int* ptr;
Foo() : ptr(new int(0)) {}
// It is valid since it does not change the state of Foo.
// It does not change where ptr points to. It just changes
// the value of what ptr points to.
void set(int value) const { *ptr = value; }
};
The compiler gives you the basic const characteristics. Higher level const-ness needs to be implemented by classes themselves.
In the case of Foo, if changing the value of what ptr to is deemed to change the state of Foo by the creator of Foo, then you'll have to remove const qualifier from the member function.
Marking a function as const means that you can't:
change any of the values of this's ivars
call a non-const function on this or any of its ivars
E.g. it would treat int foo; as const int foo;
These properties aren't transitive to pointers. Factory *_factory; doesn't become const Factory *_factory;, it becomes Factory *const _factory;, which is what you already have.
Imagine if you were using a smart pointer instead, would you expect the compiler to know that it should convert std::shared_ptr<Factory> _factory; into std::shared_ptr<const Factory> _factory;? All it would do is treat it as const std::shared_ptr<Factory> _factory;
I've got a const method in my class, which cannot be changed to non-const. In this method, I need to call a non-const method but the compiler doesn't let me do that.
Is there any way around it? Here is a simplified sample of my code:
int SomeClass::someMethod() const {
QColor saveColor = color();
setColor(QColor(255,255,255)); // Calling non-const method
// ....
setColor(saveColor); // restore color
return 1;
}
You could use const_cast on this pointer,
int SomeClass::someMethod() const {
const_cast<SomeClass*>( this )->setColor(...);// Calling non-const method
//whatever
}
but if you do that for an object that was originally declared const you run into undefined behavior.
So this:
SomeClass object;
object.someMethod();
is okay, but this:
const SomeClass object;
object.someMethod();
yields undefined behavior.
The real solution is that your const function should not be const in the first place.
One of the challenges of doing const-correctness is you can't do it halfway. It's either all or nothing. If you try to do it halfway, you end up in a tough spot like you are here. You end up with a nice const-correct class being used by some crazy old, typically legacy (or written by an old curmudgeon) code that isn't const-correct and it just doesn't work. You're left wondering if const-correctness is worth all the trouble.
I need to call a non-const method [from a const method]
You can't -- not directly. Nor should you. However, there is an alternative...
Obviously you can't call a non-const method from a const method. Otherwise, const would have no meaning when applied to member functions.
A const member function can change member variables marked mutable, but you've indicated that this is not possible in your case.
You could attempt to cast away constness by doing something like SomeClass* me = const_cast<SomeClass*>(this); but A) This will typically result in UB, or 2) It violates the whole idea of const-correctness.
One thing you could do, if what you're really trying to accomplish would support this, is to create a non-const proxy object, and do nonconst-y stuff with that. To wit:
#include <iostream>
#include <string>
using namespace std;
class Gizmo
{
public:
Gizmo() : n_(42) {};
void Foo() const;
void Bar() { cout << "Bar() : " << n_ << "\n"; }
void SetN(int n) { n_ = n; };
int GetN() const { return n_; }
private:
int n_;
};
void Gizmo::Foo() const
{
// we want to do non-const'y things, so create a proxy...
Gizmo proxy(*this);
int save_n = proxy.GetN();
proxy.SetN(save_n + 1);
proxy.Bar();
proxy.SetN(save_n);
}
int main()
{
Gizmo gizmo;
gizmo.Foo();
}
If you require to change some internal state inside a const-method you can also declare the affected state mutable:
class Foo {
public:
void doStuff() const { bar = 5; }
private:
mutable int bar;
};
This is intended for cases where you have stuff like mutexes as members of your class. Acquiring and releasing a mutex does not affect client-visible state, but is technically forbidden in a const-method. The solution is to mark the mutex mutable. Your case looks similar, although I think your class requires some refactoring for this solution to be applicable.
Also, you might want to read this answer to see how you can make this temporary state-change exception-safe using RAII.
How to call a non-const method from a const method?
You should not. You might run into undefined behaviour if you cast away the const-ness of this, using const_cast. The usage ofconst_cast will shut the compiler's mouth up, but that isn't a solution. If you need to do, then it means the const function should not be const in the first place. Make it non-const.
Or, you should do something else, which would not require you to call non-const function from const function. Like, don't call setColor function? Like, split the const function into more than one functions (if you can do that)? Or something else?
In your particular case, if setColor only sets some member variable, say m_color, then you can declare it mutable:
mutable QColor m_color;
and then set it in your const function, without calling setColor function, and without doing const_cast.
I have an a class which is singleton as defined follows
class myData {
private:
myData (void); // singleton class.
// Copy and assignment is prohibted.
myData (const myData &);
myData & operator=(const myData &);
static myData* s_pInstance;
public:
~myData (void);
static const myData & Instance();
static void Terminate();
void myFunc() { cout << "my function..." ;}
};
// In cpp file.
myData* myData::s_pInstance(NULL);
myData::myData(){}
myData::~myData()
{
s_pInstance = NULL;
}
const myData& myData::Instance()
{
if (s_pInstance == NULL)
{
s_pInstance = new myData();
}
return *(s_pInstance); // want to avoid pointer as user may deallocate it, so i used const referense
}
void main() {
(myData::Instance()).myFunc();
}
I am getting following error
error C2662: 'myData::myFunc' : cannot convert 'this' pointer from 'const myData' to 'myData&'
how to avoid above problem and call a function from Instance function which is returning const reference?
Thanks!
You'd want to declare func() as a constant member function, so the compiler knows it won't violate the const'd return value from the instance() function.
You could instead also make the instance() function return a 'regular' reference as apposed to a const one.
So either turn:
void myFunc() into void myFunc() const
Or turn:
const myData& myData::Instance() into myData& myData::Instance()
If you are calling a function on a const reference, the function you call must also be const, in your case void myFunc() const.
Otherwise you might return a non-const reference, if that works better.
The error says that myData::Instance() is a const instance of the class, and it can't call myFunc() on that, because myFunc() might change the instance, and you can't change a const instance.
Of course, you know that myFunc() can't really change the instance, but you must advertise this fact, as follows:
void myFunc() const { cout << "my function..." ;}
Avoiding the whole discussion of whether Singleton is a good to have pattern or the source of all evil, if you are actually implementing a singleton, chances are that const correctness will not work there as you expect it, so you should be aware of some pitfalls.
First your error: your Instance() static member returns a const reference, and that means that you can only perform operations that do not modify the object, i.e. call member functions marked as const, or use public members if present in a way that do not modify their values. My suggested solution is modify Instance() to return a non-const reference, rather than making func() const as others suggest.
Now for a longer explanation to the problem of const-correctness in general when applied to your particular Singleton problem. The basic problem is that when you implement a type, you divide those members that modify the object from those that don't, and you mark the latter as const member functions so that the compiler knows of your promise (allows you to call that method on a constant object) and helps you not break it (complains if you try to modify the state in the definition of the method). A method that is marked as const can be applied to both a constant and non constant object, but a method that is not marked const can only be applied to an object that is not const.
Back to the original piece of code, if you implement a singleton and the only way of accessing the object is by an Instance() method that returns a const reference, you are basically limiting all user code to use only const methods implemented in your interface. That means that effectively either all methods are non-mutating, or they are useless (const_cast should never be used). That in turn means that if you have any non-const operation you want to provide an Instance() method that returns a non-const reference.
You could consider implementing two variants of Instance(), but that will not be really helpful. Overload resolution will not help in user code to determine which version to use, so you will end up having to different methods: Instance(), ConstInstance() (choose your names), which means that it is up to user code to determine which one to use. The small advantage is that in their code, the choice of accessor will help documenting their intended usage, and maybe even catch some errors, but more often than not they will just call the non-const version because it works.