Meaning of "const" last in a C++ method declaration?
What is the meaning of const
in declarations like these? The const
confuses me.
class foobar
{
public:
operator int () const;
const char* foo() const;
};
When you add the const
keyword to a method the this
pointer will essentially become a pointer to const
object, and you cannot therefore change any member data. (Unless you use mutable
, more on that later).
The const
keyword is part of the functions signature which means that you can implement two similar methods, one which is called when the object is const
, and one that isn't.
#include <iostream>
class MyClass
{
private:
int counter;
public:
void Foo()
{
std::cout << "Foo" << std::endl;
}
void Foo() const
{
std::cout << "Foo const" << std::endl;
}
};
int main()
{
MyClass cc;
const MyClass& ccc = cc;
cc.Foo();
ccc.Foo();
}
This will output
Foo
Foo const
In the non-const method you can change the instance members, which you cannot do in the const
version. If you change the method declaration in the above example to the code below you will get some errors.
void Foo()
{
counter++; //this works
std::cout << "Foo" << std::endl;
}
void Foo() const
{
counter++; //this will not compile
std::cout << "Foo const" << std::endl;
}
This is not completely true, because you can mark a member as mutable
and a const
method can then change it. It's mostly used for internal counters and stuff. The solution for that would be the below code.
#include <iostream>
class MyClass
{
private:
mutable int counter;
public:
MyClass() : counter(0) {}
void Foo()
{
counter++;
std::cout << "Foo" << std::endl;
}
void Foo() const
{
counter++;
std::cout << "Foo const" << std::endl;
}
int GetInvocations() const
{
return counter;
}
};
int main(void)
{
MyClass cc;
const MyClass& ccc = cc;
cc.Foo();
ccc.Foo();
std::cout << "The MyClass instance has been invoked " << ccc.GetInvocations() << " times" << endl;
}
which would output
Foo
Foo const
The MyClass instance has been invoked 2 times
The const means that the method promises not to alter any members of the class. You'd be able to execute the object's members that are so marked, even if the object itself were marked const
:
const foobar fb;
fb.foo();
would be legal.
See How many and which are the uses of “const” in C++? for more information.
The const
qualifier means that the methods can be called on any value of foobar
. The difference comes when you consider calling a non-const method on a const object. Consider if your foobar
type had the following extra method declaration:
class foobar {
...
const char* bar();
}
The method bar()
is non-const and can only be accessed from non-const values.
void func1(const foobar& fb1, foobar& fb2) {
const char* v1 = fb1.bar(); // won't compile
const char* v2 = fb2.bar(); // works
}
The idea behind const
though is to mark methods which will not alter the internal state of the class. This is a powerful concept but is not actually enforceable in C++. It's more of a promise than a guarantee. And one that is often broken and easily broken.
foobar& fbNonConst = const_cast<foobar&>(fb1);
链接地址: http://www.djcxy.com/p/1054.html