问题
What is the meaning of const in declarations like these? The const confuses me.
class foobar
{
public:
operator int () const;
const char* foo() const;
};
回答1:
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 << "Foo has been invoked " << ccc.GetInvocations() << " times" << endl;
}
which would output
Foo
Foo const
Foo has been invoked 2 times
回答2:
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.
回答3:
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);
回答4:
These const mean that compiler will Error if the method 'with const' changes internal data.
class A
{
public:
A():member_()
{
}
int hashGetter() const
{
state_ = 1;
return member_;
}
int goodGetter() const
{
return member_;
}
int getter() const
{
//member_ = 2; // error
return member_;
}
int badGetter()
{
return member_;
}
private:
mutable int state_;
int member_;
};
The test
int main()
{
const A a1;
a1.badGetter(); // doesn't work
a1.goodGetter(); // works
a1.hashGetter(); // works
A a2;
a2.badGetter(); // works
a2.goodGetter(); // works
a2.hashGetter(); // works
}
Read this for more information
回答5:
Blair's answer is on the mark.
However note that there is a mutable qualifier which may be added to a class's data members. Any member so marked can be modified in a const method without violating the const contract.
You might want to use this (for example) if you want an object to remember how many times a particular method is called, whilst not affecting the "logical" constness of that method.
回答6:
Meaning of a Const Member Function in C++ Common Knowledge: Essential Intermediate Programming gives a clear explanation:
The type of the this pointer in a non-const member function of a class X is X * const. That is, it’s a constant pointer to a non-constant X (see Const Pointers and Pointers to Const [7, 21]). Because the object to which this refers is not const, it can be modified. The type of this in a const member function of a class X is const X * const. That is, it’s a constant pointer to a constant X. Because the object to which this refers is const, it cannot be modified. That’s the difference between const and non-const member functions.
So in your code:
class foobar
{
public:
operator int () const;
const char* foo() const;
};
You can think it as this:
class foobar
{
public:
operator int (const foobar * const this) const;
const char* foo(const foobar * const this) const;
};
回答7:
when you use const in the method signature (like your said: const char* foo() const;) you are telling the compiler that memory pointed to by this can't be changed by this method (which is foo here).
回答8:
I would like to add the following point.
You can also make it a const & and const &&
So,
struct s{
void val1() const {
// *this is const here. Hence this function cannot modify any member of *this
}
void val2() const & {
// *this is const& here
}
void val3() const && {
// The object calling this function should be const rvalue only.
}
void val4() && {
// The object calling this function should be rvalue reference only.
}
};
int main(){
s a;
a.val1(); //okay
a.val2(); //okay
// a.val3() not okay, a is not rvalue will be okay if called like
std::move(a).val3(); // okay, move makes it a rvalue
}
Feel free to improve the answer. I am no expert
回答9:
The const keyword used with the function declaration specifies that it is a const member function and it will not be able to change the data members of the object.
来源：https://stackoverflow.com/questions/751681/meaning-of-const-last-in-a-function-declaration-of-a-class