Need to work correctly for negative numbers

[Alb13G]

Id like to be able to have the for loop in the function to work correctly for negative numbers in either input position. Write a function that accepts two input parameters, computes their product(the result of multiplication),and returns the result. You are not allowed to use the multiplication operator(*)for performing multiplication You must implement this as repeated addition in a loop, building up the answer in an accumulator variable.
Any help will be appreciated

#include<iostream>
using namespace std;


int addition(int, int);


int main()
{
	int num1,	//to store first number
		num2,	//to store second number
		product = 0;	//to store multiplication

	   //read numbers
	cout << "Enter first number: ";
	cin >> num1;
	cout << "Enter second number: ";
	cin >> num2;

	//call function
	product = addition(num1, num2);

	//print multiplication
	cout << "The product of " << num1 << " and " << num2 << " is: " << product << endl << endl;

	system("pause");
	return 0;
}

//function definition
int addition(int a, int b)

{
	int result = 0;
	
	for (int counter = 0; counter < b; ++counter)
	{
		result = result + a;
	}
	return (result);
}





  

[Peter87]

So you have a function that calculates the product called addition?

Well, you don't need to use a loop for that. Instead you can use the * operator.

return a * b;

This will work for both positive and negative values of a and b.

[Alb13G]

Sorry I should have made this clearer mybad Write a function that accepts two input parameters, computes their product(the result of multiplication),and returns the result. You are not allowed to use the multiplication operator(*)for performing multiplication You must implement this as repeated addition in a loop, building up the answer in an accumulator variable.

[jonnin]

you have to figure out the true sign of a.
a little truth table:
a, b -> +
-a, b -> -
a,-b -> -
-a,-b ->+

if its +, do nothing. if its -, change a: a = -a;
and then change the loop a little:
for (int counter = 0; counter < abs(b); ++counter)

that should do it.

[Alb13G]

I get the for loop but I don't understand the rest is there a if/else in there or where would this fit in ?

if (a = +) do nothing

else if (a = -) a = -a ? how do you write this ?

[JLBorges]

#include <iostream>

long long product( int a, int b )
{
    // zero multiplied by anything yields zero
    if( a==0 || b==0 ) return 0 ;

    // product of two negative numbers is positive
    else if( a<0 && b<0 ) return product( -a, -b ) ;

    // product of a negative number and a positive number is negative
    else if( a<0 ) return -product( -a, b ) ;
    else if( b<0 ) return -product( a, -b ) ;

    else // both a and b are positive
    {
        // make the larger number a and the smaller number b (swap them if needed)
        // (to reduce the number of iterations)
        if( a < b ) { const int temp = a ; a = b ; b = temp ; }

        // you must implement this as repeated addition in a loop
        long long result = a ;
        for( ; b > 1 ; --b ) result += a ;
        return result ;
    }
}

int main()
{
    for( int a : { -1000, -23, -1, 0, +1, +23, +1000 } )
    {
        for( int b : { -1000, -23, -1, 0, +1, +23, +1000 } )
        {
            std::cout << a << " * " << b << " == "
                      << product(a,b) << " == " << a*b << '\n' ;
        }
    }
}

http://coliru.stacked-crooked.com/a/b39551f8ca49f9af

[jonnin]

I will look at doing the repeated sum reduction, but the quick brute force version I came up with was this. ^^ I did not flip the args if B > A (yet) as I want to figure out the reduction logic first.

int p(int a, int b)
{
  int r = 0;  
  int a2 = abs(a);
  int b2 = abs(b);  
  if( (a < 0 && b > 0) || (a>0 && b < 0))
	  a2 = -a2;
  if(a&&b)
  {
	for(int i = 0; i < b2; i++)
	  r += a2;
  }
  return r;	
}

[jonnin]

The fast version. It can be tweaked, but anyone have a better core approach?
what it does is create a bit shifter, i, which is 1,2,4,8,... etc powers of 2.
it then keeps a running addition of b. that is, when i is 1, its b. when i = 2, its b+b. When i = 4, its b+b+b+b. and so on. the sum you want is the bits of b related to that sum, eg if b were 5, which in binary is 4+1, you want (b+b+b+b) + b. Or in the code, it adds the zeros too (faster than checking), so its (b+b+b+b) +0(2's bit is zero)+ b
just like doing integer powers.

long long p(long long a, long long b)
{
  long long r = 0;  
  long long zero = 0;  
  long long a2 =  abs(a);
  long long b2 = abs(b);   
  if( (a < 0 && b > 0) || (a>0 && b < 0))
	  a2 = -a2;  
  long long* t[2] = {&a2, &zero};
    for(long long i = 1; i <= 9223372036854775808ull && b2 >= i; i= i << 1)   	
	{		
	  r += t[((unsigned long long)b2&(unsigned long long)i)==0][0]; 
	  a2+=a2;	  
	}
  return r;	
}

int main()
{
   cout << p(50000,-50000)<< endl;  //does less than 64 additions.  I think 16 for this example
   cout << p(3,-5)<< endl;
   cout << p(-3,5)<< endl;
   cout << p(3,5)<< endl;
   cout << p(0,-5)<< endl;
   cout << p(4,0)<< endl;
}

See if you can understand this. I don't advise turning it in, but if you can see how to do this, you will be way ahead of the class.

[Alb13G]

Thanks jonnin but between you and JLBorges I got it to work! Thanks to both of you for all your help its very much appreciated !

[jonnin]

Good! But do you see how 50000*50000 can be done in only 16 additions? I am not always very good at explaining things.

[Alb13G]

Thanks its a little bit beyond my scope now but I understand the beginning recursion.6 x 1 = 6
6 x 2 = 6 + (6 x 1)
6 x 3 = 6 + (6 x 2) = 6 + [6 + (6 x 1)] = 6 + 6 + 6
Wait till you see my exponent program without multiplication or pow

[jonnin]

exponent works just like addition in terms of relating the bits of the power to the answer. Its almost the exact same code.
2 to the 5th power... break 5 into its bits, 4 and 1 bit are set, so its 2*1 (1s bit) * 2*2*0(2's bit is zero) * 2*2*2*2 (4's bit) ... exact same idea.

[dhayden]

Using recursion to handle negative numbers and adding appropriate values of a*2ⁿ for multiplying positive numbers. This is similar to JLBorges's and Jonnin's solutions.

#include <iostream>

int product(int a, int b)
{
    if (a < 0) return -product(-a, b);
    else if (b < 0) return -product(a, -b);
    else {			// both are positive
	int result=0;
	for (unsigned mask=1; mask; mask <<= 1) {
	    if (b & mask) {
		result += a;
	    }
	    a <<= 1;
	}
	return result;
    }
}

int main()
{
    int a,b;
    while (std::cin >> a >> b) {
	std::cout << a << " * " << b << " = " << product(a,b) << '\n';
    }
}