Re: vector of structs?

From:
"Heinz Ozwirk" <hozwirk.SPAM@arcor.de>
Newsgroups:
microsoft.public.vc.stl
Date:
Thu, 27 Apr 2006 23:39:38 +0200
Message-ID:
<44513a15$0$11065$9b4e6d93@newsread4.arcor-online.net>
"Brian Muth" <bmuth@mvps.org> schrieb im Newsbeitrag =
news:u5pIJyjaGHA.4972@TK2MSFTNGP03.phx.gbl...

 
"Jason S" <jmsachs@gmail.com> wrote in message
news:1146169753.282126.207280@y43g2000cwc.googlegroups.com...

I've used map<T> and vector<T> and such before, where T is a simple
type, but am rather confused about the recommended way of doing it =

with

structures, especially ones that contain objects as a member. (Like
CComBSTR and CComPtr)

If I have

typedef struct {
 short s;
 MyClass obj;
 CComPtr<something> p;
} X_t;

should I use vector<X_t> or vector<X_t*>? If the former, how do I
handle push_back() with a structure? If the latter, then don't I have
to allocate memory for X_t, in which case, what's the point of using
vector<>? And if I do choose to allocate memory for X_t, I can't just
use malloc/CoTaskMemAlloc(sizeof(X_t)*N), because then how will it
properly initialize the non-simple types contained in X_t like obj()
and p() which might have constructors?


 
If you use vector<X_t *>, then you need to add code to free the =

structs

before the variable goes out of scope. I tend to chose vect<X_t> for =

that

reason (unless the struct is VERY large).
 
However, you need to then add a copy constructor and assignment =

operator.

Brush off your C++ language book and look up those terms.
 
More explicitly, you need to define:
 
X_t::X_t(const X_t &x)
 
and
 
inside your struct, implement
void operator=(const X_t &);


Usually there is no need to implement copy-ctors and assignment =
operators for struct's. In most cases the compiler will create the =
proper ones automatically. You only need you own copy-ctors, dtors and =
assignment operators if your struct (or class) contains members of =
types, that don't have their own "smart enough" copy_ctors etc, like raw =
pointers or HANDLEs. In such situations, however, it is better to use =
members with proper copy-ctors etc, instead of implementing them for the =
struct:

    // raw pointers
    struct Pointers
    {
        char* firstName;
        char* lastName;
        Pointers(): firstName(0), lastName(0) {}
        Pointers(Pointers const& x)
        {
            firstName = new char[strlen(x.firstName) + 1]; =
strcpy(firstName, x.firstName);
            lastName = new char[strlen(x.lastName) + 1]; =
strcpy(lastName, x.lastName);
        }
        ~Pointers() { delete[] firstName; delete[] lastName; }
        Pointers& operator=(Pointers const& x)
        {
            Pointers t(x);
            std::swap(*this, t);
            return *this;
        }
    };

    // strings
    struct Strings
    {
        std::string firstName;
        std::string lastName;
    };

It should be obvious which code is easier to write. And consider what =
you have to do (and what you can forget to do) when you add a middleName =
member to those struct's...

HTH
    Heinz

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