为什么必须在运算符重载中提供关键字const

Question

Just curious on why a param has to be a const in operation overloading

CVector& CVector::operator= (const CVector& param)
{
  x=param.x;
  y=param.y;
  return *this;
}

couldn't you have easily done something like this ??

CVector& CVector::operator= (CVector& param) //no const
{
  x=param.x;
  y=param.y;
  return *this;
}

Isn't when something becomes a const, it is unchangeable for the remainder of the applications life ?? How does this differ in operation overloading ???

Answer 1

You don't need const:

@numerical25: Just curious on why a param has to be a const in operation overloading

It's not required, but it is a good design decision.

See the C++ standard Section 12.8-9:

A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X& or const volatile X&

---

I think it's a good idea though:

Using a const parameter does seems like a logical design decision to me though because you want to ensure that the other value will not be changed.

It tells other people that use your class that you will not be changing the other value when you say something like: myObject = other; and it enforces this so you can't accidentally change other .

Also if you allowed non const references to the object as the parameter, then you are limiting the amount of objects that can use your function. If it is const it can be used for parameters that are const and non const. If your parameter is non const it can only be used by parameters that are non const.

---

const only applies to the current reference, not the object:

@numerical25: Isn't when something becomes a const, it is unchangeable for the remainder of the applications life ?? How does this differ in operation overloading ???

A const reference is simply that a reference that is const. It does not change the const-ness of the actual object you are passing in.

---

An example of non-const operator overloading:

Here is an example of operator overloading where the parameter is not const.
I DO NOT RECOMMEND TO DO THIS THOUGH:

class B
{
public: 
 const B& operator=(B& other)
 {
  other.x = 3;
  x = other.x;
  return *this;
 }

 int x;
};


void main(int argc, char** argv[])
{
 B a;
 a.x = 33;
 B b;
 b.x = 44;
 a = b;//both a and b will be changed
 return 0;
}

Answer 2

A const parameter is const throughout the function using it, it does not change its constness outside of it.

In this case you want to declare a const argument so that your assignment operator accepts both non-const variables and const variables; the latter case, in particular, includes the result of expressions, which is a temporary const variable and which you generally want to support in assignments.

Answer 3

If you used

CVector& CVector::operator= (CVector& param) // no const

then did this:

const CVector& my_vector = GetMyVector();
some_other_vector = my_vector; // call assignment operator - error!

You'll get an error because my_vector is a const CVector& and that can't be cast to a CVector& (non-const reference). It's just the local reference to it inside the operator= function that is const, not the entire object itself.

Answer 4

You can use the non-const variety, but this has two repercussions, one which is functional, and one which is about what you, as the writer of the function, are telling the user.

1) people calling the function that takes a non-const reference would not be able to call it using a const variable

2) when you have a function argument that's a non-const reference, you're signalling, "I reserver the right to change this". Typically, when a user of your function writes a = b;, he doesn't expect b to change.

Note that there's a third option you could use for this, pass-by-value:

CVector& CVector::operator= (CVector param) //no reference

This doesn't have either of the problems I mention above. However, it's very inefficient. Because of these three factors, passing by reference-to-const is preferred, especially in cases like a vector where copying can be expensive.

Answer 5

出于同样的原因，你可以在任何地方使用const：确保将来对方法的更改不会无意中修改传入的参数，以帮助记录接口通知调用者传递param是安全的，而不会有更改的风险，并且允许调用者传入在调用代码中声明为const的引用。

Answer 6

Another reason is to allow for conversions. For example:

string s = "foo";
s = "bar";

Here, an implementation might choose to only provide the assignment operator that takes a const reference to a string as a parameter, and depend on the compiler using a constructor to create a temporary string from the char * "bar". This would not work if the op='s parameter was not const, as you cannot bind a temporary to a non-const reference.

Answer 7

The const qualifier makes the passed parameter (in your example it is 'const CVector& param') as read only. The const qualifier ensures that the parameter (param) is not altered inside the operator=() method.

Without the const qualifier, the following is possible:

CVector& CVector::operator= (CVector& param)
{
  x=param.x;
  y=param.y;

  param.x = 10; // some random value
  param.y = 100;

  return *this;
}

The above method alters the right hand side operand 'param' after assigning the value to the left hand side operand. The const qualifier helps you not to violate the semantics of the assignment operation.