In article <JWLRi.11899$WX3.90@newsfe5-win.ntli.net>,
Phil Endecott <spam_from_usenet_0606@chezphil.org> wrote:
Hi Neelesh, thanks for the quick reply.
Neelesh Bodas wrote:
On Oct 18, 9:19 pm, Phil Endecott <spam_from_usenet_0...@chezphil.org>
wrote:
Dear Experts,
I'm surprised to find that std::max doesn't work (i.e. won't
compile) if
the arguments are not of exactly the same type, e.g. one is a short
and
the other is a long:
thats because the template is defined to take two arguments of same
type
Indeed, but I'm surpised that the short isn't promoted to a long as
it would be for a non-template function where both arguments have the
same type:
int f(long x, long y) {
return 1;
}
int g() {
short s;
long l;
return f(s,l);
}
I'm not saying that anything is wrong - this just wasn't what I had
(naively) expected.
Yes you can, just explicitly provide template arguments:
std::max<long>(s,l);
Ah, that's interesting. So if I provide an explicit type then it
behaves like my non-template function f above.
If you use it this be forewarned that you should not catch the return
value as a const reference (const long&), but catch it as a long
instead. If you catch it as a reference:
const long& m = std::max<long>(s,l);
then you risk having a reference to a destructed temporary (the
temporary long created from converting from s).