C decimal to binary without arrays

Question

I think I've almost got it, but I feel like I'm go in circles trying to figure this out. The challenge to out cout without using strings or arrays. I took the number 56 as an example and 56 should equal 111000 this is not the case as it goes through fine till 7 then the number equals number*2 + number%2 makes it equal to 15 and outputs all 1's. Idk anymore, this is driving me to the moon and back.

#include <iostream>

using namespace std;

int main()
{
int number = 0;
int n = 1;
int x = n;
cin>>number;
cout<<n%2;
while(n <= number)
{
    if(n%2 == 0)
    {
        n = n*2;
        cout<<0;
    }
    else
    {
        n = n*2 + n%2;
        cout<<n%2;
    }
}
}

DXsmiley · Accepted Answer · 2013-05-18 06:31:37Z

2

You can use the binary operator & to check if a single bit is 1 or 0.

for (int i=512; i>0; i/=2) {
    cout << ( ( number & i ) != 0 ) ;
}

Note that this WILL print leading 0's. Also, I'm assuming you only want to print positive integers.

Alternative:

for (int i=512; i>0; i/=2) {
    if (number >= i) {
        cout << 1;
        number -= i;
    } else {
        count << 0;
    }
}

edited May 18, 2013 at 6:31

answered May 18, 2013 at 6:05

DXsmiley

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5 Comments

patros Over a year ago

Good solution, but no need to call pow every time. You can start with i = 2^64 and divide by 2 (shift by one) every iteration.

victorsh Over a year ago

I can't use math functions either, no pow :(

DXsmiley Over a year ago

I've updated the solution with patros's suggestion and added an alternative method.

Brad Over a year ago

You can simply use cout << (bool)(number & i); aswell

victorsh Over a year ago

ok I see what you did there, clever :D. Thanks a lot dxsmiley and patros I would have been stuck with this for quite some time.

pakupo · Accepted Answer · 2013-05-18 07:10:14Z

0

You can use recursion

void decimal_to_binary(int decimal)
{
    int remainder = decimal % 2;
    if (decimal < 1)
        return;
    decimal_to_binary(decimal / 2);
    cout << remainder;
}

This function will take the decimal, get its remainder when divided to 2. Before it the function call itself again, it checks if the decimal is less than 1(probably 0) and return to execute the printing of 1's and 0's

answered May 18, 2013 at 7:10

pakupo

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1 Comment

victorsh Over a year ago

OOOO this is cool and quite advanced for what where I'm at. I'll hold hold onto this for later, thank you.

user2852191 · Accepted Answer · 2013-10-06 17:10:06Z

I had this type of problem assigned to me recently. This code example work up to a maximum of 10 binary digits (per the problem guidelines) and keep prompting for input until 0 is entered (sentinel value). This can certainly be improved but the math is correct:

#include <iostream>
#include <cmath>
using namespace std;

int main ()
{
 //Declare Variables
 int inputValue = 0;
 int workingValue = 0;
 int conversionSum = 0;

 //Begin Loop
 do{
     //Prompt for input
     cout << "Enter a binary integer (0 to quit): ";
     cin >> inputValue;

     //Reset Variables
     workingValue = inputValue;
     conversionSum = 0;

    //Begin processing input
    //10 digits max, so 10 iterations

    for (int i=0; i<10; i++) {
        //Check for non-binary entry
        if ((workingValue % 10) != 1 && (workingValue % 10 != 0)){
            cout << "Invalid!\n";
            workingValue = 0;
            conversionSum = 0;
            break;
           }

        //check to see if 2^i should be added to sum
        if (workingValue%2 == 1){
            conversionSum += pow(2,i);
            workingValue--;
           }
        //divide by 10 and continue loop
        workingValue= workingValue / 10;
    }

    //output results
    cout << "converted to decimal is: " << conversionSum << endl;

 }while (inputValue != 0);
}

R Balasubramanian · Accepted Answer · 2018-08-06 15:08:08Z

0

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    cout << "enter a number";
    int number, n, a=0;
    cin >> number;
    n = number;
    do 
    {
        n=n/2;
        a=a+1;  
    }
    while (n>=1);
    cout << "a is" << a;
    int c = a;
    int b = a;
    cout << "binary is";
    for(int i=0; i<=c; i++)
    {
        int k = number / pow(2,b);
        cout << k;
        number = number - k * pow(2,b);
        b = b-1;
    }
    return 0;
}

Although asked in C I have used C++. I have used the logic that if you have to convert decimal to binary we have to find the maximum power of 2 contained in the number which when added by 1 becomes the number of digit of required binary .. leftmost digit is the number of highest available power of 2 (ex in 8 highest power of 2 is 3 and 1 such is available)...then subtract this from the number and (ex 8-8=0)and search for number of next highest available power of 2 and so on.

edited Aug 6, 2018 at 15:08

R Balasubramanian

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answered Aug 1, 2018 at 18:03

amit

1

1 Comment

R Balasubramanian Over a year ago

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

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