c++: how does memberwise assignment work?

Question

Possible Duplicate:
How are C array members handled in copy control functions?

If I don't override the operator = of a class, it will use default memberwise assignment.

But what does it mean?

struct A {
    int array[100];
};
A a;
A b=a;

No error. How does b copes a'sarray? Normally array_b = array_a is invalid.

Another exampe:

struct A {
    vector<int> vec;
};
A a;
A b=a;

How does b copes a'svec? Through assignment(vec_b = vec_a), constructor(vec_b = vector<int>(vec_a)) or other mystery way?

possible duplicate of http://stackoverflow.com/questions/4164279/how-are-c-array-members-handled-in-copy-control-functions — Invictus, May 05 '12 at 12:14

Alok Save · Accepted Answer · 2012-05-05T12:12:08.990

9

A b=a;

Is not assignment, it is called as Copy Initialization.

The implicitly generated copy constructor is called to create an new object b from the existing object a.
The implicitly generated copy constructor makes a copy of the array member.

For completeness I am going to add here the standard citation from the marked duplicate.

C++03 Standard: 12.8 (Copying class objects)

Each subobject is copied in the manner appropriate to its type:

if the subobject is of class type, the copy constructor for the class is used;

if the subobject is an array, each element is copied, in the manner appropriate to the element type;

if the subobject is of scalar type, the built-in assignment operator is used.

edited May 05 '12 at 12:12

answered May 05 '12 at 11:52

Alok Save

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Generally, yes. What is a shallow copy of an array though? – Konrad Rudolph May 05 '12 at 11:55
@KonradRudolph: In this case an `memcpy`.The *shallow copy* would apply to pointers and that is the reason for **Rule of Three**. – Alok Save May 05 '12 at 11:57
@Als Word. And this also can be applied to the OP's "Normally array_b = array_a is invalid." where you can't use an assignment like that, but you can construct new arrays that way (as long as the rhs is an array literal). – Mr Lister May 05 '12 at 11:59
@ChrisA. It is if the array elements are pointers. – Mr Lister May 05 '12 at 12:00
@Als Do you mean that its a shallow copy here and modifying array variable for object a will modify the content for array variable for object b ? – Invictus May 05 '12 at 12:00
@Als I am individually able to modify array variable for object a and b without 11 affecting the content of the other – Invictus May 05 '12 at 12:02
@Ritesh on which type of objects a and b? – Mr Lister May 05 '12 at 12:03
@MrLister yeah ... i did something like this to make a check:- – Invictus May 05 '12 at 12:04
for(int i=0;i<10;i++) { cout << a.array[i]; } cout < – Invictus May 05 '12 at 12:05
The contents of the object get memcpy'd. Like Mr Fooz says, that includes non-pointer arrays. – Mr Lister May 05 '12 at 12:11
@FredOverflow here its the subobject is an array, each element is assigned, in the manner appropriate to the element type; does that mean its equivalent to deep copy ? – Ritesh just – Invictus May 05 '12 at 12:13
@Ritesh: Each element is *copied* not *assigned* there is no assignment here, this is copy construction. – Alok Save May 05 '12 at 12:16
@Als I know that. Your answer didn’t say it though, even though the question was explicitly after array members. – Konrad Rudolph May 05 '12 at 15:15
@KonradRudolph: Hmm, It was a hint I couldn't follow, I guess that's my bad :) – Alok Save May 05 '12 at 15:18

score 1 · Answer 2 · answered May 05 '12 at 11:54

1

If the members have copy constructors, they get invoked. If not, the default copy constructor does the equivalent of memcpy. See Memberwise Assignment and Initialization.

In the case of non-pointer arrays, each element is copied.

answered May 05 '12 at 11:54

Mr Fooz

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c++: how does memberwise assignment work?

2 Answers2