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muliplying without using a * operator!

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**muliplying without using a * operator!**

Aug 24th, 2005

Hello Everyone,

This is my first day here. I've not browsed the site yet. But I wanted to know the answer of these questions. I'd be glad if anyone of you can help me.

1.Write a C++ program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;
2.Exchange two numbers without using a temporary variable.
3.Find if the given number is a power of 2.
4.Multiply x by 7 without using multiplication (*) operator.

Last edited by sara.rythm; Aug 24th, 2005 at 10:32 pm. Reason: spelling mistake

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

1. could be a trick question ...

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C++ Syntax (Toggle Plain Text)

  std::cout << "numbers from 1 to 100 and 100 to 1";

2. assuming we are talking integers, use a macro like ...

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C++ Syntax (Toggle Plain Text)

// macro to swap two arguments of any type
#define SWAP(x, y)   ((x) ^= (y) ^= (x) ^= (y))

Edit: Eating crow again. Sorry, change the "of any type" to "integers only"!

3. assuming we are talking integers, whittle the number down with a series of odd integers (1,3,5,7,...) until a zero match is reached ...

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
#include <iostream>
 
using namespace std;
 
void isPowerOf2(int x)
{
    int y;
 
    y = 1;   // start of odd number sequence 1, 3, 5, 7 ...
    while(x > -1)
    {
        cout << x << endl;   // test
        x = x - y;   // subtracting a sequence of odd numbers
        y = y + 2;   // next odd number
        if (x == 0)
        {
            cout << "Bingo, we have a power of 2 number" << endl;
            break;
        }
        else if (x < 0)
            cout << "Gee, this number fails the power of 2 test" << endl;
    }
}     
 
int main()
{
    isPowerOf2(98);
    isPowerOf2(81);
    isPowerOf2(1);
    isPowerOf2(-8);  // test
 
    cin.get();  // wait
    return EXIT_SUCCESS;
}
#include <iostream>

using namespace std;

void isPowerOf2(int x)
{
    int y;
    
    y = 1;   // start of odd number sequence 1, 3, 5, 7 ...
    while(x > -1)
    {
        cout << x << endl;   // test
        x = x - y;   // subtracting a sequence of odd numbers
        y = y + 2;   // next odd number
        if (x == 0)
        {
            cout << "Bingo, we have a power of 2 number" << endl;
            break;
        }
        else if (x < 0)
            cout << "Gee, this number fails the power of 2 test" << endl;
    }
}     

int main()
{
    isPowerOf2(98);
    isPowerOf2(81);
    isPowerOf2(1);
    isPowerOf2(-8);  // test
           
    cin.get();  // wait
    return EXIT_SUCCESS;
}

Edit: put that in here because the tone went splenetic later in this thread.

4. Multiple addition will do ...

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C++ Syntax (Toggle Plain Text)

  y = x+x+x+x+x+x+x;

... as long as you don't multiply by a float!

May 'the Google' be with you!

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long time no c

**Re: muliplying without using a * operator!**

Aug 25th, 2005

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Originally Posted by vegaseat

2. use a macro like ...
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C++ Syntax (Toggle Plain Text)
// macro to swap two arguments of any type
#define SWAP(x, y)   ((x) ^= (y) ^= (x) ^= (y))
// macro to swap two arguments of any type #define SWAP(x, y) ((x) ^= (y) ^= (x) ^= (y))

That won't work for any type.
http://www.eskimo.com/~scs/C-faq/q10.3.html

"One of the methods used by statists to destroy capitalism consists in establishing controls that tie a given industry hand and foot, making it unable to solve its problems, then declaring that freedom has failed and stronger controls are necessary." --Ayn Rand

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

Ouch, I have forgotten how meticulous C is. Sorry, change the "of any type" to "integers only".

May I kindly suggest you start learning Python, then a simple ...

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  a, b = b, a

... will do. Where a and b is any objet.

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Drowzee

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

the answer to number 3 (assuming the number is an integer) is to use the modulous operator, %.

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
if(!(numbername%2))
   cout<<numbername<<" is a power of 2."<<endl;
if(!(numbername%2))
   cout<<numbername<<" is a power of 2."<<endl;

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

Thanks a lot everyone. It was quite quick. This forum seems to be good. That helped a lot. Thanks !!

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

Now you got me started, for swapping either integers or floats this will work too ...

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
#include <iostream>
 
using namespace std;
 
int main()
{
    int a = 1;
    int b = 7;
 
    cout << "a = " << a << "   b = " << b << endl; 
    // swap a with b
    a = a + b;
    b = a - b;
    a = a - b;
 
    cout << "a = " << a << "   b = " << b << endl; 
 
    cin.get();   // wait
    return EXIT_SUCCESS;
}
#include <iostream>

using namespace std;

int main()
{
    int a = 1;
    int b = 7;
    
    cout << "a = " << a << "   b = " << b << endl; 
    // swap a with b
    a = a + b;
    b = a - b;
    a = a - b;
    
    cout << "a = " << a << "   b = " << b << endl; 
    
    cin.get();   // wait
    return EXIT_SUCCESS;
}

Edit: Just an interrogative alternate, this swap is actually about five times slower then the usual temp variable swap.

May 'the Google' be with you!

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Still Learning

**Re: muliplying without using a * operator!**

Aug 25th, 2005

•

Originally Posted by vegaseat

Now you got me started, for swapping either integers or floats this will work too ...

Starting with your example, you could use a c++ template.

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
#include <iostream>
#include <limits.h>
using namespace std;
 
template<class T>
swapm(T &a,T &b)
{
	a = a + b;
	b = a - b;
	a = a - b;
};
 
int main()
{
    int a = 1;
    int b = 7;
 
    cout << "a = " << a << "   b = " << b << endl; 
	swapm(a,b);    
    cout << "a = " << a << "   b = " << b << endl; 
 
 
	float c = 1.23F;
	float d = 2.34F;
 
    cout << "c = " << c << "   d = " << d << endl; 
	swapm(c,d);    
    cout << "c = " << c << "   d = " << d << endl; 
 
    cin.get();   // wait
    return 0;
}
#include <iostream>
#include <limits.h>
using namespace std;

template<class T>
swapm(T &a,T &b)
{
	a = a + b;
	b = a - b;
	a = a - b;
};

int main()
{
    int a = 1;
    int b = 7;
    
    cout << "a = " << a << "   b = " << b << endl; 
	swapm(a,b);    
    cout << "a = " << a << "   b = " << b << endl; 

    
	float c = 1.23F;
	float d = 2.34F;
	
    cout << "c = " << c << "   d = " << d << endl; 
	swapm(c,d);    
    cout << "c = " << c << "   d = " << d << endl; 

    cin.get();   // wait
    return 0;
}

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**Re: muliplying without using a * operator!**

Aug 25th, 2005

•

Originally Posted by Drowzee

the answer to number 3 (assuming the number is an integer) is to use the modulous operator, %.
Help with Code Tags

C++ Syntax (Toggle Plain Text)
if(!(numbername%2))
   cout<<numbername<<" is a power of 2."<<endl;
if(!(numbername%2)) cout<<numbername<<" is a power of 2."<<endl;

Um, no. This only tells if the number is a multiple of 2. For power of 2, that'd be (!(x & (x - 1))).

As for #1:

First of all, "if" is not a loop. I don't know what got this idea into your head. But let's not use 'if' anyway.

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
#include <iostream>
 
void cout_nums_and_back(int low, int high) {
    (high < low)  ||  (
        std::cout << low << std::endl,
        cout_nums_and_back(low + 1, high),
        std::cout << low << std::endl
    );
    return;
}
 
int main() {
	cout_nums_and_back(1, 100);
	return 0;
}
#include <iostream>

void cout_nums_and_back(int low, int high) {
    (high < low)  ||  (
        std::cout << low << std::endl,
        cout_nums_and_back(low + 1, high),
        std::cout << low << std::endl
    );
    return;
}

int main() {
	cout_nums_and_back(1, 100);
	return 0;
}

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**Re: muliplying without using a * operator!**

#10

Aug 25th, 2005

And as for #4:

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
y = x + (x << 1) + (x << 2);
y = x + (x << 1) + (x << 2);

also works.

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