Why C++ copy constructor must use const object?

问题

I understand that when we define a class copy constructor of the class is necessary as Rule of three states. I also notice that the argument of the copy constructor is usually const as the following codes illustrate:

class ABC {
public:
    int a;
    int b;
    ABC(const ABC &other)
    { 
        a = other.a;
        b = other.b;
    }
}

My question is what would happen if the argument of the copy constructor is not const:

class ABC
{
    public:
    int a;
    int b;
    ABC(ABC &other)
    { 
        a = other.a;
        b = other.b;
    }
}

I understand that in some cases if the argument of the copy constructor is const then the second implementation would fail. Also if the argument of the copy constructor is const then the object to be copied will not change its content during the process. However, I do notice that some people still use the second implementation rather than the first one. Are there any reasons that the second implementation is preferred?

回答1:

Logically, it should make no sense to modify an object of which you just want to make a copy, though sometimes it may have some sense, like a situation where you'd like to store the number of time this object has been copied. But this could work with a mutable member variable that stores this information, and can be modified even for a const object (and the second point will justify this approach)
You would like to be able to create copy of const objects. But if you're not passing your argument with a const qualifier, then you can't create copies of const objects...
You couldn't create copies from temporary reference, because temporary objects are rvalue, and can't be bound to reference to non-const. For a more detailed explanation, I suggest Herb Sutter's article on the matter

回答2:

The last thing any consumer of your class would expect is a copy constructor that changed the object that was copied! Therefore, you should always mark as const.

回答3:

There are two reasons that const may be needed here:

It ensures that you don't accidentally "damage" the original when making the copy - this is a good thing, because you don't really want your original object to be changed when making a copy of it!
You can pass in something other than a basic object - since the constructor takes a reference, if it's not an object itself - say for example an expression.

To exemplify the second case:

 class ABC
    {
       public:
           int a;
           int b;
       ABC(const ABC &other)
       { 
         a = other.a;
         b = other.b;
       }
       ABC operator+(const ABC &other)
       {
           ABC res;
           res.a = a + other.a;
           res.b = b + other.b;
           return res;
       }
    }

  ...
  ABC A;
  a.a = 1;
  a.b = 2;
  ABC B(a+a);

This won't compile if the constructor is ABC(ABC &other), since a+a is a temporary object of type ABC. But if it's ABC(const ABC &other), we can use the temporary result of a calculation and still pass it in as a reference.

回答4:

As several other answers point out, a copy constructor that modified its argument would be an unpleasant surprise. That is not, however, the only problem. Copy constructors are sometimes used with arguments that are temporaries. (Example: return from function.) And non-const references to temporaries don't fly, as explained elsewhere on SO.

回答5:

If the copy constructor doesn't specify it's parameter as const then this fragment would not compile.

const ABC foo;
ABC bar(foo);

回答6:

Copy constructors should not modify the object it is copying from which is why the const is preferred on the other parameter. Both will work, but the const is preferred because it clearly states that the object passed in should not be modified by the function.

const is for the user only. It doesn't exist for the actual executable.

回答7:

It's not a "must" in technical sense. We even have such beast right in the standard, though it got deprecated. Follow the links for the reasoning.

The semantics of copy we expect is to leave the "template" unchanged and provide a clone that is an exact equivalent in all regards, that you have hard time to tell form original.

That being expected you shall think twice to have a copy ctor that does otherwise. It will surprise users and likely introduce bugs. And frustration and noise, just try to google for 'vector of auto_ptr' just to see the count.

The remainder of the question could be "I swear not to touch the original in implementation, but want a different signature". What signature then? Let's try T and T&.

T drops out as it would require copy ctor to be usable and we're implementing just that. Recursion: see recursion.

That leaves T&. That would actually work for a deal of cases. But just fail if your original object happens to sit around in a const form, or be temporary. Why hinder that sensible case for no rain at all?

回答8:

In addition to the fundamental assumption that copy constructors should not modify the source instance, this article elaborates on the actual technical reason for using const:

http://www.geeksforgeeks.org/copy-constructor-argument-const/

Namely that and I quote:

"... compiler created temporary objects cannot be bound to non-const references ..."

来源：https://stackoverflow.com/questions/16956693/why-c-copy-constructor-must-use-const-object

标签

c++

copy-constructor