为什么引用不能在 C++ 中重新安装

Question

C++ references have two properties:

• They always point to the same object.
• They can not be 0.

Pointers are the opposite:

• They can point to different objects.
• They can be 0.

Why is there no "non-nullable, reseatable reference or pointer" in C++? I can't think of a good reason why references shouldn't be reseatable.

Edit: The question comes up often because I usually use references when I want to make sure that an "association" (I'm avoiding the words "reference" or "pointer" here) is never invalid.

I don't think I ever thought "great that this ref always refers to the same object". If references were reseatable, one could still get the current behavior like this:

int i = 3;
int& const j = i;

This is already legal C++, but meaningless.

I restate my question like this: "What was the rationale behind the 'a reference is the object' design? Why was it considered useful to have references always be the same object, instead of only when declared as const?"

Cheers, Felix

Answer 1

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

It is not possible to change what a reference refers to after initialization. That is, once a C++ reference is initialized it cannot be made to refer to a different object later; it cannot be re-bound. 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++.

Answer 2

In C++, it is often said that "the reference is the object". In one sense, it is true: though references are handled as pointers when the source code is compiled, the reference is intended to signify an object that is not copied when a function is called. Since references are not directly addressable (for example, references have no address, & returns the address of the object), it would not semantically make sense to reassign them. Moreover, C++ already has pointers, which handles the semantics of re-setting.

Answer 3

Because then you'd have no reseatable type which can not be 0. Unless, you included 3 types of references/pointers. Which would just complicate the language for very little gain (And then why not add the 4th type too? Non-reseatable reference which can be 0?)

A better question may be, why would you want references to be reseatable? If they were, that would make them less useful in a lot of situations. It would make it harder for the compiler to do alias analysis.

It seems that the main reason references in Java or C# are reseatable is because they do the work of pointers. They point to objects. They are not aliases for an object.

What should the effect of the following be?

int i = 42;
int& j = i;
j = 43;

In C++ today, with non-reseatable references, it is simple. j is an alias for i, and i ends up with the value 43.

If references had been reseatable, then the third line would bind the reference j to a different value. It would no longer alias i, but instead the integer literal 43 (which isn't valid, of course). Or perhaps a simpler (or at least syntactically valid) example:

int i = 42;
int k = 43;
int& j = i;
j = k;

With reseatable references. j would point to k after evaluating this code. With C++'s non-reseatable references, j still points to i, and i is assigned the value 43.

Making references reseatable changes the semantics of the language. The reference can no longer be an alias for another variable. Instead it becomes a separate type of value, with its own assignment operator. And then one of the most common usages of references would be impossible. And nothing would be gained in exchange. The newly gained functionality for references already existed in the form of pointers. So now we'd have two ways to do the same thing, and no way to do what references in the current C++ language do.

Answer 4

Intrestingly, many answers here are a bit fuzzy or even beside the point (eg it's not because references cannot be zero or similar, in fact, you can easily construct an example where a reference is zero).

The real reason why re-setting a reference is not possible is rather simple.

• Pointers enable you to do two things: To change the value behind the pointer (either through the -> or the * operator), and to change the pointer itself (direct assign = ). Example:

   int a; int * p = &a;

  1. Changing the value requires dereferencing: *p = 42;
  2. Changing the pointer: p = 0;

• References allow you to only change the value. Why? Since there is no other syntax to express the re-set. Example:

   int a = 10; int b = 20; int & r = a; r = b; // re-set r to b, or set a to 20?

In other words, it would be ambiguous if you were allowed to re-set a reference. It makes even more sense when passing by reference:

void foo(int & r)
{
    int b = 20;
    r = b; // re-set r to a? or set a to 20?
}
void main()
{
    int a = 10;
    foo(a);
}

Hope that helps :-)

Answer 5

A reference is not a pointer, it may be implemented as a pointer in the background, but its core concept is not equivalent to a pointer. A reference should be looked at like it *is* the object it is referring to. Therefore you cannot change it, and it cannot be NULL.

A pointer is simply a variable that holds a memory address. The pointer itself has a memory address of its own, and inside that memory address it holds another memory address that it is said to point to. A reference is not the same, it does not have an address of its own, and hence it cannot be changed to "hold" another address.

I think the parashift C++ FAQ on references says it best:

Important note: Even though a reference is often implemented using an address in the underlying assembly language, please do not think of a reference as a funny looking pointer to an object. A reference is the object. It is not a pointer to the object, nor a copy of the object. It is the object.

and again in FAQ 8.5 :

Unlike a pointer, once a reference is bound to an object, it can not be "reseated" to another object. The reference itself isn't an object (it has no identity; taking the address of a reference gives you the address of the referent; remember: the reference is its referent).

Answer 6

A reseatable reference would be functionally identical to a pointer.

Concerning nullability: you cannot guarantee that such a "reseatable reference" is non-NULL at compile time, so any such test would have to take place at runtime. You could achieve this yourself by writing a smart pointer-style class template that throws an exception when initialised or assigned NULL:

struct null_pointer_exception { ... };

template<typename T>
struct non_null_pointer {
    // No default ctor as it could only sensibly produce a NULL pointer
    non_null_pointer(T* p) : _p(p) { die_if_null(); }
    non_null_pointer(non_null_pointer const& nnp) : _p(nnp._p) {}
    non_null_pointer& operator=(T* p) { _p = p; die_if_null(); }
    non_null_pointer& operator=(non_null_pointer const& nnp) { _p = nnp._p; }

    T& operator*() { return *_p; }
    T const& operator*() const { return *_p; }
    T* operator->() { return _p; }

    // Allow implicit conversion to T* for convenience
    operator T*() const { return _p; }

    // You also need to implement operators for +, -, +=, -=, ++, --

private:
    T* _p;
    void die_if_null() const {
        if (!_p) { throw null_pointer_exception(); }
    }
};

This might be useful on occasion -- a function taking a non_null_pointer<int> parameter certainly communicates more information to the caller than does a function taking int* .

Answer 7

It would probably have been less confusing to name C++ references "aliases"? As others have mentioned, references in C++ should be though of the variable they refer to, not as a pointer/ reference to the variable. As such, I can't think of a good reason they should be resettable.

when dealing with pointers, it often makes sense allowing null as a value (and otherwise, you probably want a reference instead). If you specifically want to disallow holding null, you could always code your own smart pointer type ;)

Answer 8

C++ references can sometimes be forced to be 0 with some compilers (it's just a bad idea to do so*, and it violates the standard*).

int &x = *((int*)0); // Illegal but some compilers accept it

EDIT: according to various people who know the standard much better than myself, the above code produces "undefined behavior". In at least some versions of GCC and Visual Studio, I've seen this do the expected thing: the equivalent of setting a pointer to NULL (and causes a NULL pointer exception when accessed).

Answer 9

This is not actually an answer, but a workaround for this limitation.

Basically, when you try to "rebind" a reference you are actually trying to use the same name to refer to a new value in the following context. In C++, this can be achieve by introducing a block scope.

In jalf's example

int i = 42;
int k = 43;
int& j = i;
//change i, or change j?
j = k;

if you want to change i, write it as above. However, if you want to change the meaning of j to mean k , you can do this:

int i = 42;
int k = 43;
int& j = i;
//change i, or change j?
//change j!
{
    int& j = k;
    //do what ever with j's new meaning
}

Answer 10

You can't do this:

int theInt = 0;
int& refToTheInt = theInt;

int otherInt = 42;
refToTheInt = otherInt;

...for the same reason why secondInt and firstInt don't have the same value here:

int firstInt = 1;
int secondInt = 2;
secondInt = firstInt;
firstInt = 3;

assert( firstInt != secondInt );

Answer 11

I agree with the accepted answer. But for constness, they behave much like pointers though.

struct A{
    int y;
    int& x;
     A():y(0),x(y){}
};

int main(){
  A a;
  const A& ar=a;
  ar.x++;
}

works. See

Design reasons for the behavior of reference members of classes passed by const reference

Answer 12

There's a workaround if you want a member variable that's a reference and you want to be able to rebind it. While I find it useful and reliable, note that it uses some (very weak) assumptions on memory layout. It's up to you to decide whether it's within your coding standards.

#include <iostream>

struct Field_a_t
{
    int& a_;
    Field_a_t(int& a)
        : a_(a) {}
    Field_a_t& operator=(int& a)
    {
        // a_.~int(); // do this if you have a non-trivial destructor
        new(this)Field_a_t(a);
    }
};

struct MyType : Field_a_t
{
    char c_;
    MyType(int& a, char c)
        : Field_a_t(a)
        , c_(c) {}
};

int main()
{
    int i = 1;
    int j = 2;
    MyType x(i, 'x');
    std::cout << x.a_;
    x.a_ = 3;
    std::cout << i;
    ((Field_a_t&)x) = j;
    std::cout << x.a_;
    x.a_ = 4;
    std::cout << j;
}

This is not very efficient as you need a separate type for each reassignable reference field and make them base classes; also, there's a weak assumption here that a class having a single reference type won't have a __vfptr or any other type_id -related field that could potentially destroy runtime bindings of MyType. All the compilers I know satisfy that condition (and it would make little sense not doing so).

Answer 13

I would imagine that it is related to optimization.

Static optimization is much easier when you can know unambiguously what bit of memory a variable means. Pointers break this condition and re-setable reference would too.

Answer 14

Because sometimes things should not be re-pointable. (Eg, the reference to a Singleton.)

Because it's great in a function to know that your argument can't be null.

But mostly, because it allows use to have something that really is a pointer, but which acts like a local value object. C++ tries hard, to quote Stroustrup, to make class instances "do as the ints d". Passing an int by vaue is cheap, because an int fitss into a machine register. Classes are often bigger than ints, and passing them by value has significant overhead.

Being able to pass a pointer (which is often the size of an int, or maybe two ints) that "looks like" a value object allows us to write cleaner code, without the "implementation detail" of dereferences. And, along with operator overloading, it allows us to write classes use syntax similar to the syntax used with ints. In particular, it allows us to write template classes with syntax that can be equally applied to primitive, like ints, and classes (like a Complex number class).

And, with operator overloading especially, there are places were we should return an object, but again, it's much cheaper to return a pointer. Oncve again, returning a reference is our "out.

And pointers are hard. Not for you, maybe, and not to anyone that realizes a pointer is just the value of a memory address. But recalling my CS 101 class, they tripped up a number of students.

char* p = s; *p = *s; *p++ = *s++; i = ++*p;

can be confusing.

Heck, after 40 years of C, people still can't even agree if a pointer declaration should be:

char* p;

or

char *p;

Answer 15

I always wondered why they didn't make a reference assignment operator (say :=) for this.

Just to get on someone's nerves I wrote some code to change the target of a reference in a structure.

No, I do not recommend repeating my trick. It will break if ported to a sufficiently different architecture.

Answer 16

Being half serious: IMHO to make them little more different from pointers ;) You know that you can write:

MyClass & c = *new MyClass();

If you could also later write:

c = *new MyClass("other")

would it make sense to have any references alongside with pointers?

MyClass * a =  new MyClass();
MyClass & b = *new MyClass();
a =  new MyClass("other");
b = *new MyClass("another");

Answer 17

C++ 中的引用不可为空的事实是它们只是别名的副作用。