Question

convert a positive integer number in C++ (0 to 2,147,483,647) to a 32 bit binary and display.

I want do it in traditional “mathematical” way (rather than use bitset or use vector *.pushback* or recursive function or some thing special in C++…), (one reason is so that you can implement it in different languages, well maybe)

So I go ahead and implement a simple program like this:

#include <iostream>
using namespace std;
int main()
{
    int dec,rem,i=1,sum=0;
    cout << "Enter the decimal to be converted: ";
    cin>>dec;
    do
    {
        rem=dec%2;
        sum=sum + (i*rem);
        dec=dec/2;
        i=i*10;
    } while(dec>0);

    cout <<"The binary of the given number is: " << sum << endl;

    system("pause");
    return 0;
}

Problem is when you input a large number such as 9999, result will be a negative or some weird number because sum is integer and it can’t handle more than its max range, so you know that a 32 bit binary will have 32 digits so is it too big for any number type in C++?. Any suggestions here and about display 32 bit number as question required?

Answer 1

Using a for-loop, and a predefined array of zero-chars:

#include <iostream>
using namespace std;
int main()
{
    int dec;
    cout << "Enter the decimal to be converted: ";
    cin >> dec;

    char bin32[]  = "00000000000000000000000000000000";
    for (int pos = 31; pos >= 0; --pos)
    {
        if (dec % 2) 
            bin32[pos] = '1';
        dec /= 2;
    }

    cout << "The binary of the given number is: " << bin32 << endl;
}

For performance reasons, you may prematurely suspend the for loop:

    for (int pos = 31; pos >= 0 && dec; --pos)

Note, that in C++, you can treat an integer as a boolean – everything != 0 is considered true.