为什么我们需要引用成员的复制构造函数

Question

I don't understand why we need a copy constructor (to begin with..) especially for the sake of reference members.

Given a class

class Foo {
   Obj &obj;
};

I must now write a copy constructor and = operator (and maybe 5 other secret things), otherwise it won't compile. Why? Why would a compiler be able to figure out that it should copy an int or string but not a reference? Same for a pointer? What does it think I might want to do instead? I have to write out exactly what the compiler writes out for every other property. What if I add, or worse, someone else adds, a new property to the class and forgets to add that to the copy ctor and operator?

Answer 1

The short answer is: because C++ standard says so. But this is not a very interesting answer.

To have a more interesting answer, we can ask a different question: what might be the rationale for such behavior?

While I do not claim to know the canonical answer to this, I would not be surprised if it is just to reduce confusion. References as members could behave unexpectedly, and copying classes with them could yield results you are not expecting.

Technically speaking, nothing stops compiler from honestly copying the referenced member from source to destination. But a lot of people would expect the class instead to "repoint" the reference (similar to pointers). But references can not be repointed, and rationale for that is given in "Design and Evolution of C++":

The reason that C++ does not allow you to rebind references is given in Stroustrup's "Design and Evolution of C++":

... I had in the past been bitten by Algol68 references where r1=r2 can either assign through r1 to the object referred to or assign a new reference value to r1 (re-binding r1) depending on the type of r2. I wanted to avoid such problems in C++.