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C++ reference rather than pointer

Started by coordz Feb 13, 2007 at 11:23 AM 9 replies 2.4k views

Original Post

Feb 13, 2007 at 11:23 AM

Say I've got a class myClass then I can write a function like myfunc(myClass &a) and then in the function use the reference to call a member function like a.myMemeberFunc() If I write the function like myfunc(myClass *a) then I have to call member functions like a->myMemberFunc() All well and good. Now, within a function (a member function of another class, myClass2, as it happens) I want to write something along the lines of myClass *refToA = &member where member is a member variable of myClass2 and is of type myClass. From then on in the function I have to use refToA->myMemberFunc() Here's the question: I want to be able to write refToA.myMemberFunc() as I could with the reference approach. How do I write the initial declaration and assignment of refToA so I can do this? TIA

ToohrVyk

238

ToohrVyk

Feb 13, 2007 at 11:34 AM

A a1;A& a2 = a1; // a2 references a1assert (&a1 == &a2);a2.member();

VerMan

Feb 13, 2007 at 11:35 AM

maybe you should provide a constructor that takes a "myClass" argument to initialize it:

myClass::myClass(myClass &x){    //and provide operator=() overloaded function    this = x;}myClass refToA(&member)refToA.myMemberFunc()

but... probably i'm wrong, well give it a try

Bregma

1,022

Bregma

Feb 13, 2007 at 2:10 PM

Quote:
Original post by VerMan
maybe you should provide a constructor that takes a "myClass" argument to initialize it:
myClass::myClass(myClass &x){    //and provide operator=() overloaded function    this = x;}myClass refToA(&member)refToA.myMemberFunc()
but... probably i'm wrong, well give it a try

Yes, you're wrong. In many ways.

Firstly, that's called a copy constructor. You should write operator=() in terms of the copy constructor and not the other way around, otherwise you will end up potentially dereferencing uninitialized values, or constructing values and immediately overwriting them with new values.

Zerothly, you should never assign to "this". Bad bad bad.

Minusonethly, your argument x is a reference to a myClass object, and you're trying to assign it to this which is of type pointer-to-myClass.

Nope, ToohrVyk provides the answer.

--smw

mbaitoff

Feb 14, 2007 at 2:16 AM

there is no way to initialize a class member the way ToohrVyk proposes. one should use initialization lists in constructor:

class c1 {};class c2 {  c1& c1ref;  c1(const c1& ref): c1ref(ref) {} };

ToohrVyk

238

ToohrVyk

Feb 14, 2007 at 2:31 AM

class c2 {  c1& c1ref;  c1(<strike>const</strike> c1& ref): c1ref(ref) {} };

coordz

Author

coordz

Feb 14, 2007 at 4:15 AM

Quote:
Original post by ToohrVyk
A a1;A& a2 = a1; // a2 references a1assert (&a1 == &a2);a2.member();

Does the job nicely. Thanks a lot.

mbaitoff

Feb 14, 2007 at 5:15 AM

Quote:

Original post by ToohrVyk

class c2 {  c1& c1ref;  c2(<strike>const</strike> c1& ref): c1ref(ref) {} };

why without const? your way you won't be able to construct c2 instance from a temporary object, i.e. you won't be able to:

c1 x, y;c2 z(x+y); // oopsconst c1 fun() {...};c2 z(fun()); // oops

Paulius Maruska

Feb 14, 2007 at 5:40 AM

Quote:
Original post by mbaitoff
Quote:
Original post by ToohrVyk
class c2 {  c1& c1ref;  c2(<strike>const</strike> c1& ref): c1ref(ref) {} };
why without const? your way you won't be able to construct c2 instance from a temporary object, i.e. you won't be able to:
c1 x, y;c2 z(x+y); // oopsconst c1 fun() {...};c2 z(fun()); // oops

Without const, because you shouldn't be able to construct c2 instance from a temporary. Think of it, c2 will save the reference of such temporary object in it's c1ref member. Bad things can (and will) happen.

ToohrVyk

238

ToohrVyk

Feb 14, 2007 at 9:36 AM

Quote:
Original post by mbaitoff
why without const?

The code does not compile with const, because you would be creating a non-constant reference (c2::c1ref) from a constant reference (ref), thus causing a compiler error.
The compiler has no reason to let it compile, since the policy is to make const-conversions implicit through use of const_cast. Besides, it makes things confusing, since you're passing a constant reference that the instance may modify!
Even if c2::c1ref was also constant, creating instances of objects that reference temporary values is a Bad Idea^TM, since the lifetime of the temporary value ends before the object is fully constructed — yielding a member variable that cannot be used except in the constructor.

mbaitoff

Feb 18, 2007 at 10:53 PM

you are right. must be no const at all. mixed it up - was thinking about copy constructors.

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