Converting Decimal to Binary...how to reverse?

[Cheddar]

Hi, so I have been trying for hours to figure out what to do to make this program work, but I am just not getting it. My assignment is to convert decimal to binary, octal, and hex. So far, I am still on binary because I have been hitting a problem. When I go to print the binary out, it's backwards. I cannot figure out how to reverse it. Here is my code:

#include<iostream>
using namespace std;
#include "stack.h"

int main()
{
    int num, total = 0;
    cout << "Please enter a decimal: ";
    cin >> num;
    while(num > 0)
    {
        total = num % 2;
        num /= 2;
        cout << total;
    }
    cout << endl;
    return 0;
}

I tried using a reverse function, only to realize that the sequence of numbers is not being stored as a string or integer, and that I do not know how to do that. So, if anybody could help me understand how to do this, that would be helpful.

Also, my professor gave me this class, but I have no idea how to implement it, or if I even need to for the binary portion:

#include <iostream>
using namespace std;

const int STACK_SIZE = 100;

class stack
{
private:
    int count;			// number of items in the stack
    int data[STACK_SIZE];
public:
    stack();
    ~stack();
    void push(const int item);	// push an item on the stack
    int pop();			// pop item off the stack
    int getCount() const;			// return count
};

stack::stack()	// constructor
{
	count = 0;	// zero the stack
}

stack::~stack() {}	// default destructor

void stack::push(const int item)
{
	if (count < STACK_SIZE)
	{
		data[count] = item;
		++count;
	}
	else cout << "Overflow!\n";
}

int stack::pop()
{
	if (count >0)
	{
		--count;
		return (data[count]);
	}
	else
	{
		cout << "Underflow!\n";
		return 0;
	}
}

int stack::getCount() const
{
	return count;
}

[coder777]

The stack is exatly what you need for your backward problem. It's easy:

#include<iostream>
using namespace std;
#include "stack.h"

int main()
{
    int num, total = 0;
    cout << "Please enter a decimal: ";
    cin >> num;
    stack s; // You need the stack;
    while(num > 0)
    {
        total = num % 2;
        num /= 2;
        s.push(total); // push 'total'(?) onto the stack
        //cout << total;
    }
    while(s.getCount() > 0) // Another loop to get the values from the stack / now in the correct order
    {
        cout << s.pop();
    }
    cout << endl;
    return 0;
}

[Cheddar]

Oh wow. I hadn't thought to use the getCount method at all. >_< Thank you very much. Anyways, I have run into another problem. I have changed my program so that now I have this:

#include<iostream>
using namespace std;
#include "stack.h"

int main()
{
    int decimal, remainder, choice;
    stack s;
    cout << "Please enter a decimal: ";
    cin >> decimal;
    cout << "Convert the number from decimal into:\n0 – Binary\n1 – Octal\n2 – Hexadecimal\n";
    cin >> choice;
    switch (choice)
    {
        case 0:
            while (decimal > 0)
            {
                remainder = decimal % 2;
                decimal /= 2;
                s.push(remainder);
            }
            break;
        case 1:
            while (decimal > 0)
            {
                remainder = decimal % 8;
                decimal /= 8;
                s.push(remainder);
            }
            break;
        case 2:
            while (decimal > 0)
            {
                remainder = decimal % 16;
                decimal /= 16;
                if (remainder > 9)
                        cout << char('A'  + remainder - 10);
                else
                    cout << remainder;
                s.push(remainder);
            }
            break;
        default:
            cout << "Invalid.";
    }
    while (s.getCount() > 0)
    {
        cout << s.pop();
    }
    return 0;
}

Binary and octal are both working fine, but I can't get hexadecimal to work. It comes out in reverse as well. I presume the problem comes from the fact that push doesn't work for the char data type, but I don't know how to resolve the issue. I need guidance here, please.

[zb0S216C]

Here's how I convert decimal to binary:

char Number(5);

for(int I(0); I < (sizeof(Number) * 8); ++I)
    std::cout << ((Number &(1 << I)) ? 1 : 0);

I'm not sure if it's cross-platform, though. With this code, though, you can determine if your CPU is big endian or little endian.

_Wazzak

[andywestken]

Your switch case for hex numbers should be just like the other cases, with no o/p.

Then modify the output loop so it knows that 10 -> a, 11 -> b, etc.

Andy

PS And factor your case statements out, too!!

[georgewashere]

sounds like GWC homework ~_~

Like andywestken said, here was how I implemented this as a function using switch cases :

//other code here
    while(hexStack.getSize() > 0)
    {
        //buffer being a previously declared int
        buffer = hexStack.pop();
       //if it buffer (last value popped) is greater then 9... enter the switch statement
       //to assign it a letter, else just display the number(since it won't need a letter 
       //representation
        if(buffer > 9)
        {
            switch(buffer)
            {
                case 10 : cout << "A"; break;
                case 11 : cout << "B"; break;
                case 12 : cout << "C"; break;
                case 13 : cout << "D"; break;
                case 14 : cout << "E"; break;
                default : cout << "F"; break;
            }
        }
        else
            cout << buffer;
    }
    cout << " in hexadecimal\n";
    cout << endl;
}

[andywestken]

@georgewashere - I assume your switch mechanism is assuming that the number is guaranteed to be in the range 0 - 15, given you're using the switch's default case to handle 15 -> F. If you are making this assumption, I would prob tweak the code to:

    while(hexStack.size() > 0)
    {
        buffer = hexStack.pop();

        switch(buffer)
        {
            case 10 : cout << "A"; break;
            case 11 : cout << "B"; break;
            case 12 : cout << "C"; break;
            case 13 : cout << "D"; break;
            case 14 : cout << "E"; break;
            case 15 : cout << "F"; break;
            default : cout << buffer;
        }
    }
    cout << " in hexadecimal\n";
    cout << endl;

But if you do want to protect against values outside the range 0-15, then you need either (a) handle 15 as a specific case and use the default for out of range, and also handle underflow in the else. Or (b) use my version and protect the switch in an if(0 <= buffer && buffer <=15) test.

A more compact, pretty standard form of the trusting version is:

    while(hexStack.size() > 0)
    {
        buffer = hexStack.pop();

        static const char hexDigits[] = "0123456789ABCDEF";
        cout << hexDigits[buffer];
    }
    cout << " in hexadecimal\n";
    cout << endl;

Which can be extended more easily (than the switch-based code) to handle, for example, base 36.

static const char base36Digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

http://en.wikipedia.org/wiki/Base_36

[Cheddar]

Hey, thanks to everyone who helped out. I ended up cleaning it up pretty nicely, I think. Here is my final product:

#include<iostream>
using namespace std;
#include "stack.h"

int main()
{
    int decimal, remainder, choice, base, intBuffer;
    stack s;
    cout << "Please enter a decimal: ";
    cin >> decimal;
    cout << "Convert the number from decimal into:\n0 – Binary\n1 – Octal\n2 – Hexadecimal\n";
    cin >> choice;
    if(decimal == 0)
    {
        if(choice >= 0 && choice <= 2)
            cout << 0;
        else
            cout << "Invalid.";
    }
    if(decimal != 0)
    {
        switch(choice)
        {
            case 0:
                base = 2;
                break;
            case 1:
                base = 8;
                break;
            case 2:
                base = 16;
                break;
            default:
                cout << "Invalid.";
                return 0;
        }
    }
    while(decimal > 0)
    {
        remainder = decimal % base;
        decimal /= base;
        s.push(remainder);
    }
    while(s.getCount() > 0)
    {
        intBuffer = s.pop();
        if(choice == 2 && intBuffer > 9)
            cout << char('A'  + intBuffer - 10);
        else
            cout << intBuffer;
    }
    return 0;
}

If anyone can think of how to further clean it up, I would be glad to listen.

@georgewashere - This IS GWC homework! It's Rex! Finally found a good opportunity to ask for help here. :P