let baseclass know size of object

**Victor Bazarov** · Nov 13 '08, 04:25 PM

Re: let baseclass know size of object

..rhavin grobert wrote:

guess you have a bouquet of paddingless structs (and your compiler
already cares for that) that all have one in common: their first
memeber has to be their size. As fas as i know (am i right?) a
baseclass get's in it's ctor the _baseclass-size_ when doing a sizeof
(this), so you cant fill that member in ctor automatically. Lets say
you want the derived-class way, e.g.:
>
typedef unsigned int UINT;
struct base {
base(UINT nSize) : m_nSize(nSize) {};
UINT size() const {return m_nSize;};
private:
UINT m_nSize;
};
>
struct d1, public base {

struct d1: public base {

// .... //
};
>
now what would be the best way to automatically fill the m_nSize
memeber w/o calling a special init()-member for every derived class or
doing something like:
>
struct d. public base {

struct d: public base {

d() : base(sizeof(thi s)) {};

d() : base(sizeof(*th is)) {}

};
>
>
because depending on the deriviation, ctor d() may or may not have
some parameters and additional initalisations, so a macro would in the
end look even more complicated. is there a (perhaps templates) way to
say:
>
"if the struct is derived from 'base', one member of the ctors init-
list has to be a 'base(sizeof(th is))' " ...?

Mmm... I don't know of any. Perhaps your class documentation should say
that... Besides, what if somebody wants to give your 'size' some value
determined only during run-time? You shouldn't limit them to providing
the sizeof(*this) automatically.. .

V
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**Bart van Ingen Schenau** · Nov 14 '08, 11:05 AM

Re: let baseclass know size of object

On 13 nov, 16:50, ".rhavin grobert" <cl...@yahoo.de wrote:

guess you have a bouquet of paddingless structs (and your compiler
already cares for that) that all have one in common: their first
memeber has to be their size. As fas as i know (am i right?) a
baseclass get's in it's ctor the _baseclass-size_ when doing a sizeof
(this), so you cant fill that member in ctor automatically. Lets say
you want the derived-class way, e.g.:
>

<snip>

is there a (perhaps templates) way to
say:
>
"if the struct is derived from 'base', one member of the ctors init-
list has to be a 'base(sizeof(th is))' " ...?

If the restriction that it only works for dynamically-allocated
structures is not a problem, you could define a class-specific
operator new() to collect this information.
Something like this:

<pseudocode>
class Base {
private:
static std::size_t lastSize;
public:
static void* operator new(std::size_t size)
{
lastSize = size;
return ::operator new(size);
}
public:
Base() : mSize(lastSize)
{
lastSize = 0; /* reset to avoid undetectable bogus results */
}
private:
unsigned int mSize;
};
</pseudocode>

Bart v Ingen Schenau

**.rhavin grobert** · Nov 14 '08, 05:45 PM

Re: let baseclass know size of object

On 14 Nov., 12:01, Bart van Ingen Schenau
<Bart.van.Ingen .Sche...@ict.nl wrote:

On 13 nov, 16:50, ".rhavin grobert" <cl...@yahoo.de wrote:
>

guess you have a bouquet of paddingless structs (and your compiler
already cares for that) that all have one in common: their first
memeber has to be their size. As fas as i know (am i right?) a
baseclass get's in it's ctor the _baseclass-size_ when doing a sizeof
(this), so you cant fill that member in ctor automatically. Lets say
you want the derived-class way, e.g.:

>
<snip>

is there a (perhaps templates) way to
say:

>

"if the struct is derived from 'base', one member of the ctors init-
list has to be a 'base(sizeof(th is))' " ...?

>
If the restriction that it only works for dynamically-allocated
structures is not a problem, you could define a class-specific
operator new() to collect this information.
Something like this:
>
<pseudocode>
class Base {
private:
static std::size_t lastSize;
public:
static void* operator new(std::size_t size)
{
lastSize = size;
return ::operator new(size);
}
public:
Base() : mSize(lastSize)
{
lastSize = 0; /* reset to avoid undetectable bogus results */
}
private:
unsigned int mSize;};
>
</pseudocode>
>
Bart v Ingen Schenau

nice idea, thx, i'll try!

**peter koch** · Nov 14 '08, 08:35 PM

Re: let baseclass know size of object

On 13 Nov., 16:50, ".rhavin grobert" <cl...@yahoo.de wrote:

guess you have a bouquet of paddingless structs (and your compiler
already cares for that) that all have one in common: their first
memeber has to be their size. As fas as i know (am i right?) a
baseclass get's in it's ctor the _baseclass-size_ when doing a sizeof
(this), so you cant fill that member in ctor automatically. Lets say
you want the derived-class way, e.g.:
>
typedef unsigned int UINT;
struct base {
base(UINT nSize) : m_nSize(nSize) {};
UINT size() const {return m_nSize;};
private:
UINT m_nSize;
>
};
>
struct d1, public base {
// .... //
>
};
>
now what would be the best way to automatically fill the m_nSize
memeber w/o calling a special init()-member for every derived class or
doing something like:
>
struct d. public base {
d() : base(sizeof(thi s)) {};
>
};
>
because depending on the deriviation, ctor d() may or may not have
some parameters and additional initalisations, so a macro would in the
end look even more complicated. is there a (perhaps templates) way to
say:
>
"if the struct is derived from 'base', one member of the ctors init-
list has to be a 'base(sizeof(th is))' " ...?

You could also use the CRTP, although that would mean that you would
not derive directly from bar, but from an intermediate class:

template < typename T >
class BT: public B
{
BT(): B(sizeof(T)) {}
};

class D: public BT<D>
{
....
};

/Peter

let baseclass know size of object

let baseclass know size of object

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