Re: array of reference?
On Dec 6, 5:52 pm, terminator <farid.mehr...@gmail.com> wrote:
On Dec 6, 12:41 am, "Alf P. Steinbach" <al...@start.no> wrote:
* terminator:
On Dec 5, 12:32 am, "Alf P. Steinbach" <al...@start.no> wrote:
* siddhu:
why can't we have array of references.
A reference is not an object, it has no size.
it has a size usually equal to that of a pointer in case
of usage an argument to a none inlined function.
Sorry, that's incorrect. Or not even wrong. If you want to
make an argument, cough up chapter and verse from the
standard.
I do not get it:
void f(int& iref);
can you explain how 'iref ' is passed to 'f' if it is not
inlined? The simplest way is to implement it as a pointer ,so
stack register will be decreased sizeof(int*) for passing
'iref'.
First, inlining or not is irrelevant. The standard specifically
states that inlining has no effect on a function, so it can't
possibly the size of an argument to the function can't possibly
depend on that. Second, how the compiler passes arguments with
reference type is it's business, not mine. My compiler
certainly doesn't pass them on the stack; it puts them in a
register. A 64 bit register, even when I'm compiling in 32 bit
mode (where pointers are 32 bits). And on my compiler, the
actual memory space used by a reference in a structure will
depend on what is in front of and behind the reference. (The
same is true of pointers, of course.)
What is relevant is that the compiler says that sizeof( char& )
== 1, and sizeof( T& ) == sizeof( T ), regardless of the size of
a pointer. A reference doesn't have size. At least not of its
own. A reference is an other name for a variable or a value,
and its "size" is the size of whatever it names.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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