why reture "const &"?

[zealotcat]

template <class T>
inline T const& max (T const& a, T const& b)
{
// if a < b then use b else use a
return a<b?b:a;
}

thanks very much!!

[Vladimir]

simont wrote:

You /can't/ "assign the max of a and b to 2" in any meaningful sense.
Writing
| max(5,7) = 2;
doesn't "assign the max of 5 and 7", because you can't change that.

You can't assign 5 to 2 in any meaningful sense either:

5 = 2;

Does this mean we have to get rid of assignment operator??

Writing max(a, b) = c makes perfect sense either in math or in
programming
At least for those who knows what references are used for ;)
--
Vladimir

[Dave Rahardja]

ze*******@gmail .com wrote:

template <class T>
inline T const& max (T const& a, T const& b)
{
// if a < b then use b else use a
return a<b?b:a;
}

thanks very much!!

Because the input variables are const references. You shouldn't return a
non-const reference to a const object.

By the way, why is everyone on this thread assuming that the original
poster is talking about std::max?

I do like the idea of max returning an lvalue. The expression

max(a, b) = 5;

is succint, clear, and unambiguous.

-dr

[Mike Hewson]

ze*******@gmail .com wrote:

template <class T>
inline T const& max (T const& a, T const& b)
{
// if a < b then use b else use a
return a<b?b:a;
}

thanks very much!!

The parameters when the function 'max' is called have type 'T const&' -
meaning 'a' and 'b' are T's ( whatever they are ) that cannot be altered
in this context ( the 'const' ), and reference semantics are used (
possibly to avoid pointer stuff, but there are other reasons ).
For the type 'T' one hopes that the '<' operator is satisfactorily and
sensibly defined with regard to the meaning and usage of type 'T'.
The expression 'a<b' is evaluated in terms of its equivalence to 0, so
the conditional-expression operator as applied ( 'a<b?b:a' ) will
represent 'a' if 'a<b' is non-zero ( 'true' }, and 'b' otherwise.
The function thus returns one of 'a' or 'b' according to that test, but
with the type 'T const&'.
Presumably, this is because if the 'max' function ought not alter 'a'
or 'b' ( the 'const' in the argument list suggests this intention ),
then neither should the function that called 'max' ( to which either 'a'
or 'b' will be returned ).
Assigning to max(a,b) would seem to violate that intent, that is:

max(a,b) = Whatever; // Whatever is of type 'T'.

would be illegal as it stands. Some other function named 'max', but with
a different signature, could be used however - a non-const version for
instance.

--

Cheers
--
Hewson::Mike
"This letter is longer than usual because I lack the time to make it
shorter" - Blaise Pascal