How to pass references by value in C++?

Question

I'm trying to create am immutable type (class) in C++,

I made it so that all methods "aka member functions" don't modify the object and return a new instance instead.

I'm running across a bunch of issues, but they all revolve around the reference types in C++.

One example is when passing parameters of the same class type by reference:

Imm Imm::someOp( const Imm& p_im ) const
{
   ...
   p_im = p_im.someOtherOp(); //error, p_im is const, can't modify it!
   ...
}

The error is caused by passing the value by reference. If instead, I was passing the reference by value, then the error line above would not be an error!

Consider a Java/C# example

class Imm
{
    ...
    Imm someOp( Imm p_im )
    {
        ....
        p_im = p_im.someOtherOp(); //ok, you're not modifying the 
                 //original object, just changing the local reference
        ....  
    }
    ....
}

How can I do something like this in C++? I know I can use pointers but then I run into the whole memory management mess. I don't want to worry about who owns references to objects.

Ideally I'd like to design the class to be like immutable strings in python; you can use them without ever noticing or even knowing that they're immutable, and they just behave as you expect; they just work.

EDIT

Of course I can get around it by passing-by-value or by using a temp variable (which is what I'm doing currently). What I'm asking about is "how to pass references by value in C++"

I'm expecting the answer to revolve around something in the STL, I'm currently looking into smart_ptr family of templates.

UPDATE

Thanks for the responses, I realize there's no escape from pointers. (see my other question, which is really a follow up on this one)

Answer 1

C++ has something better than immutable types—const. A single type can be mutable or not depending on your needs. That said, there are useful patterns for dealing with short-lifetime (near-)copies:

void f(const X &x) {
  // Trivial case: unconditional copy
  X x2=transform(x);
  // Less trivial: conditional copy
  std::optional<X> maybe;
  const X &use=need_copy ? maybe.emplace(transform(x)) : x;
  use.go();  // whichever, x or *maybe
}  // *maybe destroyed iff created, then x2 destroyed

std::unique_ptr may be used in a similar fashion prior to C++17, although the function can then of course throw std::bad_alloc.

Answer 2

You need to make a new copy of your incoming argument. You can do what you want in several equivalent ways: 1) you could pass-by-value:

Imm Imm::someOp( Imm im ) const {
   im = im.someOtherOp();      // local im is a copy, original im not modified
   return im;                  // return by value (another copy)
}

or 2) you can pass-by-reference and make a copy explicitly:

Imm Imm::someOp( const Imm & im ) const {
   Imm tmp = im.someOtherOp(); // local tmp is a copy
   return tmp;                 // return by value (another copy)
}

both forms are equivalent.

Answer 3

Check this to know about temporary lifetime http://herbsutter.wordpress.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/

Imm Imm::someOp( const Imm& p_im ) const
{
   ...
   //Imm& im = p_im.someOtherOp();       // will *not* work
   const Imm& im = p_im.someOtherOp();   // will work, but you get a const reference
   ...
}

But you can use boost::shared_ptr

shared_ptr<Imm> Imm::someOtherOp() const
{
  shared_ptr<Imm> ret = new Imm;
  ...
  return ret;
}

shared_ptr<Imm> Imm::someOp(const share_ptr<Imm>& p_im) const
{
  shared_ptr<Imm> im = p_im->someOtherOp();
}

Answer 4

It doesn't work like that in C++.

When you pass a reference to an object, you are actually passing the address in memory of the object. References can't be re-seated to other objects either, hence the C++ adage "the reference IS the object." You HAVE to make a copy to modify it. Java will do this behind the scenes for you. C++, you just have to copy it.

Answer 5

In Java and C#, you are not really dealing with a reference - they are more like handles or pointers. A reference in C++ is really another name for the original object, not a pointer to it (although it may be implemented with a pointer). When you assign a value to a reference, you are assigning to the object itself. There is confusion in that to initialize a reference you can use the = character, but it is an initialization, not an assignment.

 Imm im, im2, im3; 
 Imm &imr = im;  // initialize a reference to im
 imr = im2; // assign im2 to imr (changes the value of im as well)
 Imm *imp = &im; // initialize a pointer to the address of im
 imp = &im3; // assign the address of im3 to imp (im is unnaffected);
 (*imp) = im2; // assign im2 to imp (modifies im3 as well).

If you specifically want to pass "references by value" then you are essentially asking for a contradition in terms. References, by definition are passed by reference. As pointed out elsewhere, you can pass a pointer by value, or else a straight value. If you really want, you can hold onto a reference in a class and pass that around by value:

 struct ImmRef
 {
     Imm &Ref;
     ImmRef(Imm &ref) : Ref(ref) {}
 };

Note also that a const applied to a reference is making the referred to object constant, not the reference. References are always const.

Answer 6

Didn't you forget to set the method you call as const?

EDIT: so, with the const fixed.

Maybe you should do something like

Imm & tmp = p_im.someOtherOp();

Then do further operation on the tmp variable.

If you set a variable or a parameter as const & you just cannot assign to it.

Answer 7

Isn't assignment, by definition, not a constant operation?

You look like you're trying to assign something to a const reference, which totally defeats the idea of a const reference.

I think you may be looking for a pointer instead of a reference.