>>>>> On 21 Apr 1995 17:54:55 GMT, mshafiq@aol.com said:

mshafiq> Can someone please help!

I wouldn't expect *anybody* to help you in this group. At least in my
understanding this group is for discussion of standard issues. Maybe
you should try comp.lang.c++ or something like that.
--

                      -- Jens --
         http://wowbagger.pc-labor.uni-bremen.de/jens.html

In article <3nb582$l1v@warp.cris.com>, mshafiq@aol.comORshafiq@cris.com says...
>This gives me an error :
>member 'number::val' of class 'number' is not accessible.
>Obviously because 'val' can only be accessed by member functions of class
>number! So my dilemma is, how can I get an overloaded sqrt() function to
>take and return parameters of type number?

Make the overloaded function a *friend*.

--
Trent Glascock
glascock@esd.dl.nec.com

Thankyou for your suggestion(s).
However, I am still slightly confused as to how would the sqrt function
then knows to access the private variable stored inside the  number
class?

I tried making the function sqrt() a non-member function, thus
..
number sqrt(number &n2)
{
number temp;
temp.val = sqrt(n2.val);
return temp;
}

This gives me an error :
member 'number::val' of class 'number' is not accessible.
Obviously because 'val' can only be accessed by member functions of class
number! So my dilemma is, how can I get an overloaded sqrt() function to
take and return parameters of type number?

I am a student of C++ trying to overload the sqrt() function so that it
takes and returns a self-defined object of type 'number'.
I have been assigned this exercise purely to demonstrate operator and
function overloading - the overloaded operators are working fine, but
when I try to call the overloaded sqrt() function thus:

CODE FRAGMENT BELOW
..

class number {
 double val;
public:
 number(double d){val=d;};
 number() {val=0.0;};
 void get_number (double &d){d = val;};
 void operator=(double d){val = d;};
 number operator +(number n2);
        .
        .
 number sqrt(number n2);
        }

..
..
..
number number::sqrt(number n2)
{
  number temp;
  temp.val= sqrt(n2.val);
  return temp;
}

..
..
..
n3=sqrt(n2);


where n2 and n3 are objects of type 'number' with a private variable of
type double,
I receive the error "cannot convert from number to a double" (Symantec
compiler).
Can someone please help!

I am a student of C++ trying to overload the sqrt() function so that it
takes and returns a self-defined object of type 'number'.
I have been assigned this exercise purely to demonstrate operator and
function overloading - the overloaded operators are working fine, but
when I try to call the overloaded sqrt() function thus:

CODE FRAGMENT BELOW
..

class number {
 double val;
public:
 number(double d){val=d;};
 number() {val=0.0;};
 void get_number (double &d){d = val;};
 void operator=(double d){val = d;};
 number operator +(number n2);
        .
        .
 number sqrt(number n2);
        }

..
..
..
number number::sqrt(number n2)
{
  number temp;
  temp.val= sqrt(n2.val);
  return temp;
}

..
..
..
n3=sqrt(n2);


where n2 and n3 are objects of type 'number' with a private variable of
type double,
I receive the error "cannot convert from number to a double" (Symantec
compiler).
Can someone please help!

I am a student of C++ trying to overload the sqrt() function so that it
takes and returns a self-defined object of type 'number'.
I have been assigned this exercise purely to demonstrate operator and
function overloading - the overloaded operators are working fine, but
when I try to call the overloaded sqrt() function thus:

CODE FRAGMENT BELOW
..

class number {
 double val;
public:
 number(double d){val=d;};
 number() {val=0.0;};
 void get_number (double &d){d = val;};
 void operator=(double d){val = d;};
 number operator +(number n2);
        .
        .
 number sqrt(number n2);
        }

..
..
..
number number::sqrt(number n2)
{
  number temp;
  temp.val= sqrt(n2.val);
  return temp;
}

..
..
..
n3=sqrt(n2);


where n2 and n3 are objects of type 'number' with a private variable of
type double,
I receive the error "cannot convert from number to a double" (Symantec
compiler).
Can someone please help!

hello baby
love you
xxxoxxx

mshafiq@aol.com writes:

>I am a student of C++ trying to overload the sqrt() function so that it
>takes and returns a self-defined object of type 'number'.
>I have been assigned this exercise purely to demonstrate operator and
>function overloading - the overloaded operators are working fine, but
>when I try to call the overloaded sqrt() function thus:
>
>CODE FRAGMENT BELOW
>..
>
>class number {
> double val;
>public:
> number(double d){val=d;};
> number() {val=0.0;};
> void get_number (double &d){d = val;};
> void operator=(double d){val = d;};
> number operator +(number n2);
>        .
>        .
> number sqrt(number n2);
>        }
>
>..
>..
>..
>number number::sqrt(number n2)
>{
>  number temp;
>  temp.val= sqrt(n2.val);
>  return temp;
>}
>
>..
>..
>..
>n3=sqrt(n2);
>
>
>where n2 and n3 are objects of type 'number' with a private variable of
>type double,
>I receive the error "cannot convert from number to a double" (Symantec

You are declaring number::sqrt(number) as a member function, when
your usage indicates that you really want a non-member function.

Simply remove the sqrt declaration from the number class and declare
the function:

   number sqrt(number n2) { ... }

If necessary, declare this function as a friend of class number.

In article <3n8rea$146@warp.cris.com> mshafiq@aol.com writes:

   From: mshafiq@aol.com
   Newsgroups: comp.std.c++
   Date: 21 Apr 1995 17:53:14 GMT
   Organization: Concentric Research Corporation

   I am a student of C++ trying to overload the sqrt() function so that it
   takes and returns a self-defined object of type 'number'.
   I have been assigned this exercise purely to demonstrate operator and
   function overloading - the overloaded operators are working fine, but
   when I try to call the overloaded sqrt() function thus:

   CODE FRAGMENT BELOW
   ..

   class number {
    double val;
   public:
    number(double d){val=d;};
    number() {val=0.0;};
    void get_number (double &d){d = val;};
    void operator=(double d){val = d;};
    number operator +(number n2);
    .
    .
    number sqrt(number n2);
    }

   ..
   number number::sqrt(number n2)
   {
     number temp;
     temp.val= sqrt(n2.val);
     return temp;
   }

   ..
   ..
   ..
   n3=sqrt(n2);


   where n2 and n3 are objects of type 'number' with a private variable of
   type double,
   I receive the error "cannot convert from number to a double" (Symantec
   compiler).
   Can someone please help!

-------------------------------------------------


    nowhere  do you have the means to create a 'number'

    from another number.  there are doubles everywhere,

    but the compiler will not juggle them all around to

    convert things for you


    ---- at the least you need an operator=, and a copy
         ctor that take 'number' as input

     the compiler will NOT :

         take a number.....make it into a double by inventing
         a function I have not defined myself.....
         pass that double to a 'number' ctor......
         make a temp 'number'....use it to redefine
         a pre-existing 'number'.....

       the ability to flip/flop between double and number
       exists only in YOUR perception.  You only told
       the compiler one side of the story....it won't make
       up the rest of the story for you

---------------------------------