C++ - Decimal to binary converting

Question

I wrote a \'simple\' (it took me 30 minutes) program that converts decimal number to binary. I am SURE that there\'s a lot simpler way so can you show me? Here\'s the code:<

Answer 1

#include <iostream>
#include <bitset>

#define bits(x)  (std::string( \
            std::bitset<8>(x).to_string<char,std::string::traits_type, std::string::allocator_type>() ).c_str() )


int main() {

   std::cout << bits( -86 >> 1 ) << ": " << (-86 >> 1) << std::endl;

   return 0;
}

Answer 2

An int variable is not in decimal, it's in binary. What you're looking for is a binary string representation of the number, which you can get by applying a mask that filters individual bits, and then printing them:

for( int i = sizeof(value)*CHAR_BIT-1; i>=0; --i)
    cout << value & (1 << i) ? '1' : '0';

That's the solution if your question is algorithmic. If not, you should use the std::bitset class to handle this for you:

bitset< sizeof(value)*CHAR_BIT > bits( value );
cout << bits.to_string();

Answer 3

You can use std::bitset to convert a number to its binary format.

Use the following code snippet:

std::string binary = std::bitset<8>(n).to_string();

I found this on stackoverflow itself. I am attaching the link.

Answer 4

You want to do something like:

cout << "Enter a decimal number: ";
cin >> a1;
cout << setbase(2);
cout << a1

Answer 5

#include <iostream>

// x is our number to test
// pow is a power of 2 (e.g. 128, 64, 32, etc...)
int printandDecrementBit(int x, int pow)
{
    // Test whether our x is greater than some power of 2 and print the bit
    if (x >= pow)
    {
        std::cout << "1";
        // If x is greater than our power of 2, subtract the power of 2
        return x - pow;
    }
    else
    {
        std::cout << "0";
        return x;
    }
}

int main()
{
    std::cout << "Enter an integer between 0 and 255: ";
    int x;
    std::cin >> x;

    x = printandDecrementBit(x, 128);
    x = printandDecrementBit(x, 64);
    x = printandDecrementBit(x, 32);
    x = printandDecrementBit(x, 16);

    std::cout << " ";

    x = printandDecrementBit(x, 8);
    x = printandDecrementBit(x, 4);
    x = printandDecrementBit(x, 2);
    x = printandDecrementBit(x, 1);

    return 0;
}

this is a simple way to get the binary form of an int. credit to learncpp.com. im sure this could be used in different ways to get to the same point.

Answer 6

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

void Decimal2Binary(long value,char *b,int len)
{
    if(value>0)
    {
        do
        {
            if(value==1)
            {
                *(b+len-1)='1';
                break;
            }
            else
            {
                *(b+len-1)=(value%2)+48;
                value=value/2;
                len--;
            }
        }while(1);
    }
}
long Binary2Decimal(char *b,int len)
{
    int i=0;
    int j=0;
    long value=0;
    for(i=(len-1);i>=0;i--)
    {
        if(*(b+i)==49)
        {
            value+=pow(2,j);
        }
        j++;
    }
    return value;
}
int main()
{
    char data[11];//最後一個BIT要拿來當字串結尾
    long value=1023;
    memset(data,'0',sizeof(data));
    data[10]='\0';//字串結尾
    Decimal2Binary(value,data,10);
    printf("%d->%s\n",value,data);
    value=Binary2Decimal(data,10);
    printf("%s->%d",data,value);
    return 0;
}