I read here that if I don't write a copy constructor the compiler does it for me using the assignment operator, which results in shallow copy of Objects. What if I do have the assignment operator overloaded in all of my member object? wouldn't it result in a deep copy?
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3 Answers
if I don't write a copy constructor the compiler does it for me using the assignment operator
No, it doesn't use the assignment operator; it does it via a implicitly generated copy constructor which does a shallow copy.
What if I do have the assignment operator overloaded in all of my member object? wouldn't it result in a deep copy?
Given that the assignment operator is not used in absence of explicitly defined copy constructor, even though you have the assignment operator overloaded you still need to overload the copy constructor as well.
Read the Rule of Three in C++03 & Rule of Five in C++11.
answered Sep 21, 2011 at 19:35
5 Comments
Kerrek SB
These Rules of N should also come with a warning, "don't write any of them (unless you have a really good reason to).
Lightness Races in Orbit
@Kerrek: And then another one that says "ignore Kerrek SB"
Alok Save
@KerrekSB: Maybe, Just say, Don't use pointer members unless you have a very good reason to. :)
Kerrek SB
@Tomalak: If you know your tools, you'll do fine, of course. But many times the reasoning is, "C++ is hard. Oh, now I know classes. Oh, now I can put my
char * in a vector. Now how do I write the destructor." etc. Point is, with well composed class design, writing any of the Big 3/5 should be a rare, localised occurrence.Lightness Races in Orbit
@Kerrek: Sure, but we have to assume that the programmer does have a good reason, because otherwise they would not have to read the Rules in the first place. This is the fail-safe approach. Those of us who know what we are doing are free to ignore the rules that we know we don't need.
The important part in the linked article is "each member of the class individually using the assignment operator." So it doesn't matter if you define the assignment operator for you class, it will use the assignment operator for each member of your class.
answered Sep 21, 2011 at 19:36
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You're misinformed. The implicitly generated constructors and assignment operators simply perform construction or assignment recursively on all members and subobjects:
copy constructor copies element by element
move constructor moves element by element
copy assignment assigns element by element
move assign move-assigns element by element
This logic is the reason why the best design is one in which you don't write any copy constructor (or any of the other three, or the destructor) yourself, and instead compose your class of well-chosen, single-responsibility classes whose own semantics take care of everything.
answered Sep 21, 2011 at 19:38
7 Comments
Alok Save
A nitpick The Q is not tagged C++11 actually.
Kerrek SB
@Als: I could say that the question isn't tagged C++03 :-)
Alok Save
But then being pedantic C++11 has a separate tag here, So C++03 is the only one being referred here by C++ tag, it doesn't have a separate tag. ;-)
Lightness Races in Orbit
Actually I suppose we can just clarify versions as and when required, like we do for any other versioning. e.g. "I'm using jQuery 1.5.2"
Kerrek SB
@Tomalak: That's what I figured. For anything that's sufficiently version-independent, the "c++" tag is just fine. "c++11" should be only for explicitly new stuff, while the purported "c++98/03" tag could be for things that explicitly do not want to use any new features.
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