[ReMaS]

assalam 3laikom

peace

first i،m new here

and my english language is not so gooood

so please help me

i have an assignment

Write a code to Multiplication two numbers without using (+, - , * , ()) operators? ?

note : iknow this prog when useing add operator !!!

This post has been edited by ReMaS: 09 January 2010 - 12:30 PM

[FrozenSnake]

[rules][/rules]

Read this: http://www.dreaminco...topic140144.htm

And from the BIG yellow box you are suppose to read and not ignore when creating a thread: # We will not do your homework for you! Do not ask us to give you code!

This post has been edited by FrozenSnake: 09 January 2010 - 06:05 AM

[ReMaS]

FrozenSnake, on 9 Jan, 2010 - 05:04 AM, said:

[rules][/rules]

Read this: http://www.dreaminco...topic140144.htm

And from the BIG yellow box you are suppose to read and not ignore when creating a thread: # We will not do your homework for you! Do not ask us to give you code!

ok i see

make me some min to praint the code or my ideas

thanks & sorry

[JackOfAllTrades]

http://www.dreaminco...wtopic57137.htm

[NickDMax]

I am confused by the condition:

"Do not use any mathematical operators"

What operators count as "mathematical"?

I am sure +, -, *, /, and % are on the list, but what about &, |, <<, >>, ~?

are boolean operators considered "mathematical"?

what about the assignment operator =, or +=, -=, *=, /=?

what about ++, --?

Exactly what operators are forbidden?

[ReMaS]

NickDMax, on 9 Jan, 2010 - 07:31 AM, said:

I am confused by the condition:

"Do not use any mathematical operators"

What operators count as "mathematical"?

sorry

yes mr/mrs u r all rights

i was missen this type of condition


the quation is

Write a code to Multiplication two numbers without using (+, - , * , ()) operators?

just


and sorry for confused


and i tryed and this is the result i used the div operator ^_!

#include<stido.h>
void main()
{
int x=3 , y=5 , sum;
sum = x / (1 / y);
printf("%d",sum);
}

if any body have another idea >>> i.m here

and thaks for every body

[optix212]

there are actually alot of ways you could do this, but i would just create a function.

#include<stido.h>
void main()
{
int x=3 , y=5 , sum;
sum = x / (1 / y);
printf("%d",sum);
}

the "/" symbol is the mathematical operator for divide, so i think this would be wrong.

[Aphex19]

This thread might help you.

[NickDMax]

WEll I was just having some fun and thought: Well if you can add, then you can multiply... so I used the info on this page here to make a binary adder... Though my adder seems to have an error if both numbers are negative... Anyway here is my adder (note that the code in main() is just to test the adder):

#include <iostream>
#include <iomanip>
#include <ctime>
#include <cstdlib>

using namespace std;

int binaryAdder(int a, int b) {
	int carry = 0;
	int value = 0;
	int place = 0;
	unsigned int A = (unsigned)a;
	unsigned int B = (unsigned)b;
	while(A!=0 || B!=0) {
		int bita = A % 2;
		int bitb = B % 2;
		int aXORb = bita ^ bitb;
		int aANDb = bita & bitb;  
		int carryANDaXORb = carry & aXORb;
		int temp = aXORb ^ carry;
		value |= temp << place++;
		carry = aANDb | carryANDaXORb;
		A >>= 1;
		B >>= 1;
	}
	value |= carry << place;
	return value;
}
		



int main() {
	srand(time(0));
	for (int i = 0; i < 100; i++) {
		int a = rand() % 1000;// * (rand()%2 ? -1: 1);
		int b = rand() % 1000;// * (rand()%2 ? -1: 1);
		int result = binaryAdder(a, b);
		cout << setw(5) << a 
			 << " + " 
			 << setw(5) << b 
			 << " = " 
			 << setw(5) << binaryAdder(a, b) 
			 << "\t| Error = " 
			 << result - (a+b) 
			 << endl;
	}
}

[ReclusiveManiac]

ReMaS, I'm not sure why no one has commented on your code, but it is perfectly valid to multiply two numbers as long as you are allowed to use the / operator. Your code should work great.

x / (1 / y) = x * y

[NickDMax]

Quote

ReMaS, I'm not sure why no one has commented on your code, but it is perfectly valid to multiply two numbers as long as you are allowed to use the / operator. Your code should work great.

x / (1 / y) = x * y

Note that in C/C++ using integers this is NOT a true statement!!!

1/y = 0

try 1.0/y

[ReclusiveManiac]

Nevermind, typing the code I just realized you used parentheses which your assignment prohibited.

Edit: To NickDMax you are correct. You do need the 1.0 so the resulting number stays as a decimal. Good catch.

This post has been edited by ReclusiveManiac: 09 January 2010 - 01:02 PM

[NickDMax]

The restriction on () is pretty useless...

double temp = 1.0/y;
int result = x / temp;

[ReclusiveManiac]

Yeah I knew there was a way you could code it without the parentheses, but I couldn't figure it out immediately. Nice job man!

[optix212]

well i don't see how this assignment would be possible without making something along the lines of a binary adder, like Nick did. But, if the division sign is allowed, then i think this assignment would be really easy, given that division is just the opposite of multiplication.