I have this class:
class BankAccount{
private:
char* ownerName;
char IBAN[14];
double balance;
}
And I have this function:
char* BankAccount::getIban(){
return this->IBAN;
}
That one is valid but I wonder why I can't define getIban() like this, because I want to make sure that IBAN can't be changed :
char* BankAccount::getIban()const{
return this->IBAN;
}
It says return value does not match the function type.
Inside a const function all of the members behave as if they were const, in your case the member IBAN is equivalent to const char IBAN[14]. You cannot get a non-const char* to refer to a const array, and thus the error. You probably want to do:
const char* BankAccount::getIban() const {
return IBAN;
}
you have to include const in the function declaration in the header
If you declare a method const when defining it the compiler WONT LET YOU potentially ruin that const-ness by passing pointers to const stuff outside where they're not const.
The error is that. You need to return a const char*, that way the compiler knows when you use that function the resulting type is a const char* <-- you can look but not touch, so the IBAN value stays const.
If you make the const method return a const char* (return IBAN) it'll be fine, because when you use that method C++ wont let you change what the result of calling it points to.
Related
I am trying to to understand const_cast.
In the example below, items_ is a private member variable.
The method getItems() is a const method, which means that it cannot modify the member variable items_.
getItems() returns a const vector reference which means that you cannot modify the member variable items_.
My question is, is const_cast necessary here? I don't think so as I don't need to cast away constness?
#include <iostream>
class test {
std::vector<Item> items_;
public:
const std::vector<Item>& getItems() const;
};
const std::vector<mv::Item>& mv::Workloads::getItems() const
{
return const_cast<std::vector<Item>&>(items_);
}
It is not only unnecessary but it is plain wrong in that place.
Consider what happens when you remove the const from the return type:
/*const*/ std::vector<mv::Item>& mv::Workloads::getItems() const
{
return const_cast<std::vector<Item>&>(items_);
}
Now the method returns a non-const reference to a const object! Note that the method itself is declared as const, hence this is const in the context of this method and also items_ is const. Using the reference returned by that method would lead to undefined behaviour when the method is called on a const test instance.
In that sense, the only effect of the const_cast here is to potentially silence an important compiler error should you ever decide to change the return type.
On the other hand, in the method as is (ie returning a const reference) there is no reason for the cast. items_ is const and a const reference to it is returned.
const_cast is useful eg to avoid code duplication for const and non-const methods when you do know for sure that the object you cast const away really is not const, see here for details.
PS: in case you actually do want to return a mutable reference to a member from a const method, you would declare the member mutable. Also then no const_cast would be required. Though, mutable should be used with care and usually isnt needed in the first place.
It is not neccessary, simply return the _items.
Const member functions see the the object itself (*this) as const, so every data member is seen as const, from which building and returning const reference is allowed.
#include <iostream>
using namespace std;
class test {
std::vector<Item> items_;
public:
const std::vector<Item>& getItems() const;
};
const std::vector<mv::Item>& mv::Workloads::getItems() const
{
return items_;
}
I have a ClassA that has a private: vector<ClassB> vec. I'm filling the vector up in ClassA::fillVec().
Now i'd like to return the vector(by reference? so no copying) and i'd also like forbid any further changes using const.
What still confuses me is the syntax. What i have so far is
const std::vector<ClassB> &ClassA::fillVec(...) const {}
But I don't know if that is right. And even if it's right, I found this solution on the internet, so if anyone could explain why the two const
The first const means that the return type is const reference i.e. the vector may not be modified through the reference.
The const at the end means that the member function is not allowed to modify the (ClassA) object. It is therefore allowed to call that method on a const ClassA instance. This of course contradicts with the purpose of the function assuming it's supposed to modify the member; it should therefore not be const.
You want to return a const reference to prevent the user changing it; but the function itself can't be const, since it modifies a class member.
const std::vector<ClassB> &fillVec(<parameters>);
^ ^
const return value no const here
You would use the second const on member functions that aren't supposed to modify the object they're called on.
I'm still learning about C++ and I'm reading everywhere that I have to use const everywhere I can (for speed reason I think).
I'm usually write my getter method like this:
const bool isReady() {
return ready;
}
But I've seen that some IDE autogenerate getter in this way:
bool getReady() const {
return ready;
}
But, writing delegates, it happened to me to find this error if the const is after the function:
member function 'isReady' not viable: 'this' argument has type 'const VideoReader', but function is not marked const
So that, what is the better way to write a const getter? Do I really have to care about?
There is a huge difference between the two ways.
const bool isReady()
The code above will return a const bool, but it does not guarantee that the object will not change its logic state.
bool isReady() const
This will return a bool, and it guarantees that the logic state of your object will not change. In this case it is not necessary to write const in front of the return type. It makes no sense to return a const bool because it is a copy anyway. So making it const is useless. The second const is needed for const correctness, which is not used for speed reasons but to make your program more reliable and safe.
They mean two differnt things:
const bool isReady() {
return ready;
}
This returns a constant bool. Meaning a bool which cannot change value from the time it's been created.
bool getReady() const {
return ready;
}
This is a a constant function, meaning a function that will not alter any member variables of the class it belongs to. This is the style recommended to use for getters, since their only purpose is to retrieve data and should not modify anything in the process.
const method informs compiler that you will not modify class instance on which this method is called:
class A {
public:
bool getReady() const {
return ready;
}
};
so if you try to modify your object inside getReady() then compiler will issue error. Const methods are usefull where you have ie.: const A&, or const A*, then you can call only const methods on such objects.
as for:
const bool isReady() {
return ready;
}
this const provides actually no real benefit, because bool is copied while isReady() returns. Such const whould make sense if returned type was a const char* or const A&, in such cases const makes your char string or A class instance immutable.
A const getter has the signature
bool getReady() const
The other version isn't a const method, it just returns a const value (which is basically useless).
Having a const getter allows you to call it on const objects:
const Object obj;
obj.getReady();
This is only valid if getReady is marked as const.
There is a difference between using the const keyword for the return type or for the method signature. In the first case the returned value will be a constant value. In the second case the method will be a so-called constant method, which cannot change the representation of the object. On constant objects, only the constant methods are callable.
I just started learning C++, switching from a JAVA environment.
When reading through some Boost examples I found the following two methods defined in a class:
const char* data() const
{
return data_;
}
char* data()
{
return data_;
}
There are two things that confuse me.
First is the reserved word const, which I think I understand here. The first const refers to the char* which means that I cannot change the value of the pointer. The second const tells me that calling the function will not make changes to the state of the object I am calling. Is that a correct interpretation?
Second point of confusion is why one would have two methods with the same name and signature. How does the compiler know which one I meant to call? How do I know whether I am allowed to change the data after calling data() without know which of the two I called?
The first function returns a pointer to constant data. The const at the end of the function signature indicates that the function will not modify and class data members.
The second function returns a pointer to mutable data. The caller can use the pointer to modify the class member variable.
Search the web and SO for "const correctness".
There are several ways to declare const within a variable with a pointer:
const char * // you can modify the pointer, not the item pointed to
char const * // you can modify the pointer, not the item pointed to
char * const // you can modify the item pointed to, not the pointer
const char const * // you cannot modify either
char const * const // you cannot modify either
const char * const // you cannot modify either
As for the question, these two method definitions are overloads, and which one is called depends on the context. For example, if the caller is also in a const method, and making the call on one of it's members (who is an instantiation of the class that has the data() methods), then the const char * data() const method will be called, and caller can only save the return value in a variable of type const char *. Here is an example:
class MyClass {
DataClass data_obj_; // has both 'data()' methods described in your question
...
void my_method () const { // within this const method, data_obj_ cannot be modified
const char * data = data_obj_.data(); // calls const method
...
Here is an example where you would need both.
class foo {
char* data_;
public:
const char* data() const
{
return data_;
}
char* data()
{
return data_;
}
};
const foo bar;
foo bar2;
bar->data(); // this uses the const version.
bar2->data(); // this uses the non-const version.
Because bar is a const object it would be an error to call data() non-const version.
I have a program and many of its classes have some operators and methods with the keyword const like the followings:
operator const char* () const;
operator char* ();
void Save(const char *name) const;
void Load(const char *name);
First: what does it mean const at the end of the method declaration?, is it the same like putting it at the beginning?
Second: Why would be a const version and a no const version of operator() needed?
Thanks in advance.
First: what does it mean const at the end of the method declaration?, is it the same like putting it at the beginning?
No. A const at the end means that the method may be called on objects that are declared const. A const at the beginning means that the returned value is const.
Second: Why would be a const version and a no const version of operator() needed?
The non-const version returns a char* which is not const. By modifying this char* you could then in fact modify the object (assuming the char* is a member of the object).
Since this is not allowed for const objects, there's an overload of operator() for const objects, so that a const char* is returned, so the object can't be modified through it.
'const' at the end tells the compiler that this method does not change any member variables - that it is safe to call this method on const instances. So, Save could be called on a const instance, since it won't change that instance. Load on the other hand, will change the instance so can't be used on const instances.
The const version of operator() passes back a const pointer, guaranteeing the buffer passed back won't change. Presumably that's a pointer into a instance variable of the class. For non-const instances, the other operator() passes back a non-const pointer. It would have to be a pointer to some memory that even if written to, wouldn't change the contents of the instance.
Also, look up the 'mutable' keyword sometime. Understanding that will help you understand this idea of const-correctness.
Member function constness. It means the function can not(*) modify any of your member variables. It's like putting a const in front of all your member variables for this one function call. It's a good guarantee for the clients of your class and may also aid in compiler optimisations.
(*) - see also the keyword mutable.
Putting const at the end of a method declaration is stating that the object itself, or this, is const instead of the return type.
C++ allows methods to be overloaded on const for complicated reasons. Not enough space to go into full detail here. But here are a couple of short ones.
Ocassionally there is value, or flat necessity, in having a method behave differently when it is called from a const type. The most straight forward example is when you want to return a const value from a const method and a non-const value from a normal method.
Whether or not this is const dramatically changes the binding of the internal method. To the point that it would essentially become two different method bodies. Hence it makes sense to break it up into 2 different methods.
One note in addition to the other answers: there is no operator() in your example.
operator const char* () const;
operator char* ();
are conversion operators, which mean that objects of the class can be implicitly converted to C-style strings, like
void f(const MyClass& x, MyClass& y) {
const char* x_str = x;
char* y_str = y;
}
A declaration and usage of operator(), which means you can use an object of the class type sort of like a function, would look like:
class MyClass {
public:
const char* operator() (int x, int y) const;
// ...
};
void g(const MyClass& obj) {
const char* result = obj(3, 4);
}
If you're looking for a great resource on C++ (including tips on using const correctly) try "Effective C++".
A useful site about this: JRiddel.org
In C++ when you declare a method const by putting it AFTER the method signature you are asserting that "This method will not change any non-mutable instance variables in the object it is being called on."
The const before the return value (e.g. the const in: operator const char*...") is declaring that it only returns a variable pointer to a const char*. (You may not change the contents of the char* but you can re-assign the pointer.) If you wrote "const char* const ..." it would be a constant pointer to constant characters. (The const comes after the star).
The multiple versions are useful so the compiler can understand this:
const char* my_const_var = <object_name>();
char* my_var = <object_name>();
Chris
You should refer to the "HIGH·INTEGRITY C++ CODING STANDARD MANUAL" for knowing when it is recommended to use the const modifier for class members:
High Integrity CPP Rule 3.1.8: Declare 'const' any class member function that does not modify the externally visible state of the object. (QACPP 4211, 4214)
Justification: Although the language enforces bitwise const correctness, const correctness should be thought of as logical, not bitwise. A member function should be declared const if it is impossible for a client to determine whether the object has changed as a result of calling that function. The 'mutable' keyword can be used to declare member data which can be modified in const functions, this should only be used where the member data does not affect the externally visible state of the object.
class C
{
public:
const C& foo() { return * this; } // should be declared const
const int& getData() { return m_i; } // should be declared const
int bar() const { return m_mi; } // ok to declare const
private:
int m_i;
mutable int m_mi;
};
Reference Effective C++ Item 21;Industrial Strength C++ 7.13;
Const at the beginning applies to the return value. Const at the end applies to the method itself. When you declare a method as "const" you are saying that you have no intention of modifying any of the member variables of the class in the method. The compiler will even do some basic checks to make sure that the method doesn't modify member variables. The const in the return value prevents the caller from modifying the value that is returned. This can be useful when you return pointers or references to data managed by the class. This is often done to avoid returning copies of complex data which could be expensive at run time.
The reason you have two different operators is that the "const" version returns a const pointer to what is probably data internal to the class. If the instance of the class is const, then chances are you want the data being return should also be const. The "non-const" version just provides a method that returns a modifiable return value when the caller has a non-const instance of the class.