Velocity Reviews > How can we perfom multiplication programatically without using + or * operator.

# How can we perfom multiplication programatically without using + or * operator.

mjdeesh_hi@yahoo.co.in
Guest
Posts: n/a

 05-29-2006

How can we perfom multiplication programatically without using + or *
operator.
Can any one help out in this one.

santosh
Guest
Posts: n/a

 05-29-2006
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> How can we perfom multiplication programatically without using + or *
> operator.
> Can any one help out in this one.

If you conceptualise the numerical values in a binary format, then left
shifting the value by N is similar to multiplying the value by 2^N.

Richard Heathfield
Guest
Posts: n/a

 05-29-2006
(E-Mail Removed) said:

> How can we perfom multiplication programatically without using + or *
> operator.

Addition can be done by using ^ (xor), & (and), and << (left shift) in the
proper way.

Alternatively, you can double and halve, if you're careful in the case of an
odd number. Note that doubling will require using addition as described
above (since you are not allowed the * operator).

That's enough of a hint - now do your own homework.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Ico
Guest
Posts: n/a

 05-29-2006
(E-Mail Removed) wrote:
>
> How can we perfom multiplication programatically without using + or *
> operator.
> Can any one help out in this one.
>

#include <stdio.h>

int mult(int a, int b)
{
int c = 0;
while(a--) c -= b;
return -c;
}

int main(void)
{
int a = 5;
int b = 6;

printf("%d x %d = %d\n", a, b, mult(a, b));

return 0;
}

--
:wq
^X^Cy^K^X^C^C^C^C

Richard Heathfield
Guest
Posts: n/a

 05-29-2006
Ico said:

> (E-Mail Removed) wrote:
>>
>> How can we perfom multiplication programatically without using + or *
>> operator.
>> Can any one help out in this one.
>>

>
> #include <stdio.h>
>
> int mult(int a, int b)
> {
> int c = 0;
> while(a--) c -= b;
> return -c;
> }

Very funny.

But I suspect his teacher is trying to get him to find out how computers do
addition and multiplication "under the hood" (because a computer does not
have the luxury of a magic wand we call '*').

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

newbie
Guest
Posts: n/a

 05-29-2006

Richard Heathfield wrote:
> Ico said:
>
> > (E-Mail Removed) wrote:
> >>
> >> How can we perfom multiplication programatically without using + or *
> >> operator.
> >> Can any one help out in this one.
> >>

> >
> > #include <stdio.h>
> >
> > int mult(int a, int b)
> > {
> > int c = 0;
> > while(a--) c -= b;
> > return -c;
> > }

>
> Very funny.
>
> But I suspect his teacher is trying to get him to find out how computers do
> addition and multiplication "under the hood" (because a computer does not
> have the luxury of a magic wand we call '*').
>
> --
> Richard Heathfield
> "Usenet is a strange place" - dmr 29/7/1999
> http://www.cpax.org.uk
> email: rjh at above domain (but drop the www, obviously)

Hehehehehehe!

Totally irrelevent but I thought your magic wand comment was funny

santosh
Guest
Posts: n/a

 05-29-2006
newbie wrote:
> Richard Heathfield wrote:

.... snip ...

> > But I suspect his teacher is trying to get him to find out how computers do
> > addition and multiplication "under the hood" (because a computer does not
> > have the luxury of a magic wand we call '*').

>
> Hehehehehehe!
>
> Totally irrelevent but I thought your magic wand comment was funny

Have you been exposing yourself to nitrous oxide?

Richard Bos
Guest
Posts: n/a

 05-29-2006
(E-Mail Removed) wrote:

> How can we perfom multiplication programatically without using + or *
> operator.

#include <stdio.h>

int main(void)
{
double m1=4.0, m2=5.0;

printf("%f times %f equals %f.\n", m1, m2, m1/(1.0/m2));

return 0;
}

HTH; HAND.

Richard

Ico
Guest
Posts: n/a

 05-29-2006
Richard Bos <(E-Mail Removed)> wrote:
> (E-Mail Removed) wrote:
>
>> How can we perfom multiplication programatically without using + or *
>> operator.

>
> #include <stdio.h>
>
> int main(void)
> {
> double m1=4.0, m2=5.0;
>
> printf("%f times %f equals %f.\n", m1, m2, m1/(1.0/m2));
>
> return 0;
> }

Taking stupid assignments literally, that must be some kind of dutch
humor

--
:wq
^X^Cy^K^X^C^C^C^C

SM Ryan
Guest
Posts: n/a

 05-29-2006
(E-Mail Removed) wrote:
#
#
#
# How can we perfom multiplication programatically without using + or *
# operator.
# Can any one help out in this one.

Any good text on computation theory should include the multiplication

--
SM Ryan http://www.rawbw.com/~wyrmwif/
Mention something out of a Charleton Heston movie, and suddenly
everybody's a theology scholar.