On 11 Sep., 12:22, Jiang <goo.mai...@yahoo.com> wrote:
On Sep 11, 12:32 am, Martin Bonner <martinfro...@yahoo.co.uk> wrote:
sizeof(U) >= sizeof(T), but it reallistically is possible for
sizeof(U) > sizeof(T).
If T can be any type then it is possible that sizeof(U)
will be even smaller than sizeof(T).
No, that is impossible, v.i.
#include <iostream>
struct t
{
char a[32];
};
typedef t& T;
OK, T is a reference type, but this is *not* the same as T.
struct U
{
T b; // ignore its initialization ...
};
U contains now a *reference* to a t. Note also, that it is
impossible to omit the initialisation of a variable of reference type.
int main()
{
std::cout << "sizeof(T):" << sizeof(T) << std::endl;
std::cout << "sizeof(U):" << sizeof(U) << std::endl;
}
$ ./aout
sizeof(T):32
sizeof(U):4
This is expected and does not disprove Martin's assertion.
There exists a special rule in C++ for sizeof in namely [expr.sizeof]/2 :
"When applied to a reference or a reference type, the result is the size
of the referenced type[..]"
which effectively guarantees:
static_assert(sizeof(X) == sizeof(X&), "sizeof is immune to
references");
sizeof(T) > sizeof(U). (Ignore issues about initialization; they can
be fixed). I agree my initial reaction was "that's cheating", but if
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