I came across this question recently. My goal is to understand how the C++ compiler views struct definitions which hold standard library containers such as std::vector.
Ben Voigt's answer to the linked question cites the following from the C++0x standard:
....
A trivial class is a class that has a trivial default constructor (12.1) and is trivially copyable.
[ Note: In particular, a trivially copyable or trivial class does not have virtual functions or virtual base classes. — end note ]
A standard-layout class is a class that:
- has no non-static data members of type non-standard-layout class (or array of such types) or reference,
....
I'm almost certain that the bolded text implies that the following is undefined behavior
struct A
{
std::vector< SomeType > myVec;
int myC;
A( int c ) : myC : (c) {}
};
int main( void )
{
A one( 1 );
A two( 2 );
SomeType k, z;
one.myVec.push_back( k );
two.myVec.push_back( z );
memcpy( &two, &one, sizeof( A ) ); // bad juju
}
And the same would be the case for any type which is from the standard library, including simpler types such as std::string. This would be due to the nature of the library's design, given its large usage of inheritance and template programming.
So, while struct A would resemble that of a POD type, the fact that it contains that standard library type automatically invalidates it from that category, as far as the compiler is concerned.
Are my assumptions correct?
