* Alf P. Steinbach:
> * Howard Hinnant:
>> 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).
>
> Possibly you meant something other than what you actually wrote.
>
> std::max returns the same type as the argument type.

Dang, yes it does, but it adds "const&".

Sorry.

I was unable to believe something like that could have made it into the
standard.


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