Re: Template notation

From:
Salt_Peter <pj_hern@yahoo.com>
Newsgroups:
comp.lang.c++
Date:
23 Apr 2007 12:08:11 -0700
Message-ID:
<1177355291.759068.18890@n76g2000hsh.googlegroups.com>
On Apr 23, 9:13 am, desktop <f...@sss.com> wrote:

Erik Wikstr=F6m wrote:

On 23 Apr, 14:30, desktop <f...@sss.com> wrote:

I am trying to understand the following template:

template <typename T>
inline T const& max (T const& a, T const& b) {
        return a < b ? b : a;

}

does it say that the function max returns a constant reference to T of
any type?


It returns a const reference of type T, yes.

I have never seen a function returning references before, does it mean
that it returns the address of the value returned?


No, pointers are not references, a reference can be seen as an alias
for a variable, so the following is true for references but not for
pointers:

int i = 1;
int& j = i; // j is a reference to i
if (&i == &j)
  // Always true

So you can see that if you take the address of a reference to an
object you get the address of the object while if you take the address
of a pointer to a object you get that pointer's address.

Another example:

void foo(int& i) {
  ++i;
}

int k = 1;

foo(k);

Now, in the function foo(), the i we are changing is the same integer
as k, so it's another name but the same variable.

--
Erik Wikstr=F6m


Ok guess it the same as in java where Object myobj = new Object()
generates the reference myobj to an Object.


Java might call that a reference, its closer to a C++ pointer.
Java has no equivalence of a C++ reference.

The parameters to the max template:

max (T const& a, T const& b)

are also references, why pass a reference to a template instead of the
"real" object?


Its the other way around. passing by reference IS passing the "real"
variable using an alias.
Passing by value only modifies a local variable and more importantly -
it invokes a constructor, usually a copy ctor.

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