Re: std::vector and a bit-wise copy

From:
Joshua Maurice <joshuamaurice@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 9 Feb 2011 20:41:00 CST
Message-ID:
<188f51e2-12c7-43ed-9a60-a112dcbe4640@o21g2000prn.googlegroups.com>
On Feb 9, 3:24 pm, Jeremy <bjerem...@yahoo.com> wrote:

I have a question that I having trouble finding a straight answer to,
but thought it may be obvious to someone here.

If I have a struct with a vector as a member:

struct A
{
    std::vector<int> theVec;

};

And I make two instances of that vector x and y and populate the
vector in x. Is it undefined behavior to then set y= x ? Such as

int main()
{
    A x;
    A y;

    fill(x); // allocates the vector and populates it

    y = x;

}

Seeing A has no copy-constructor declared, we should do a bit-wise
copy from x to y. Since the vector in y has no memory allocated to it,
I'm assuming this should be just a case of memory trampling and
exhibit UB.

Is that wrong or am I missing something w.r.t. the bit-wise copy?


If there is no explicitly defined copy constructor or assignment
operator in a class (or struct) type, then the compiler will attempt
to provide a default one. The default is member-wise copy construction
or assignment.

In this case, you are using the compiler provided default assignment
operator "A& A::operator= (A Const& )". The default is member-wise
assignment of the members according to their "operator=". In this
case, the compiler default "A& A::operator= (A Const& )" will be as
if:
  A& A::operator= (A Const& rhs)
  {
    theVec = rhs.theVec;
    return *this;
  }

The assignment of the std::vector member is definitely not a simple
bitwise copy of its internal members, and thus there is no simple
bitwise copy in that program fragment.

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