convert decimal to 32 bit binary?

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?

Given the fact you already checked an answer I guess there's no more need for a 'mathematical' answer. — , Oct 11 '12 at 19:34

Alex Bakulin · Answer 1 · 2012-05-21T10:05:32.550

8

What you get in sum as a result is hardly usable for anything but printing. It's a decimal number which just looks like a binary.

If the decimal-binary conversion is not an end in itself, note that numbers in computer memory are already represented in binary (and it's not the property of C++), and the only thing you need is a way to print it. One of the possible ways is as follows:

int size = 0;
for (int tmp = dec; tmp; tmp >>= 1)
    size++;
for (int i = size - 1; i >= 0; --i)
    cout << ((dec >> i) & 1);

Another variant using a character array:

char repr[33] = { 0 };
int size = 0;
for (int tmp = dec; tmp; tmp >>= 1)
    size++;
for (int i = 0; i < size; ++i)
    repr[i] = ((dec >> (size - i - 1)) & 1) ? '1' : '0';
cout << repr << endl;

Note that both variants don't work if dec is negative.

edited May 21 '12 at 10:05

answered May 21 '12 at 09:55

Alex Bakulin

1,668
11
17

The fact that the value is physically represented as binary in memory isn't particularly relevant to printing the binary representation this way, just as printing it in decimal doesn't require the value to be physically represented in decimal. For example, the `>>` operator in C++ is defined in powers of two, so even on a ternary system it would produce the same values (though it would be more expensive). – bames53 May 21 '12 at 11:51
@bames53 If the number is represented, say, in ternary, then in order to print it in ternary we don't _need_ to convert it like operator `>>` does. It'd be enough to just read and print the stored digits. – Alex Bakulin May 21 '12 at 11:58
"It'd be enough to just read and print the stored digits." My point can be restated in the following way; that accessing ternary digits on a ternary machine looks just like the arithmetic you'd do on any other machine to produce the ternary representation. I.e. a bunch of divides and mods by 3. In that sense the physical representation is irrelevant. Similarly, producing a binary representation looks the same whether you're on a decimal, ternary or binary machine. – bames53 May 21 '12 at 12:21
@bames53 Now I see your point. You're right. My mistake was in thinking that on ternary machine operator `>>` would have "shift one ternary digit right" semantics. – Alex Bakulin May 21 '12 at 12:41

score 4 · Answer 2 · answered May 21 '12 at 09:34

4

You have a number and want its binary representation, i.e, a string. So, use a string, not an numeric type, to store your result.

answered May 21 '12 at 09:34

R. Martinho Fernandes

228,013
71
433
510

score 3 · Answer 3 · answered May 21 '12 at 09:32

3

You could use an unsigned integer type. However, even with a larger type you will eventually run out of space to store binary representations. You'd probably be better off storing them in a string.

answered May 21 '12 at 09:32

dirkgently

108,024
16
131
187

ok any suggestion if use string to represent 32 bit binary number? if you enter 5, binary is 101 but it have to be 29 '0' digits before 101? – Ronaldinho Learn Coding May 21 '12 at 09:36
question required to display 32 bit binary number, that's why? – Ronaldinho Learn Coding May 21 '12 at 09:40
@LongBodie: Then yes, you will need to zero-pad your string. – dirkgently May 21 '12 at 09:41

score 3 · Accepted Answer · answered May 21 '12 at 11:17

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.

score 1 · Answer 5 · answered May 21 '12 at 10:34

As others have pointed out, you need to generate the results in a string. The classic way to do this (which works for any base between 2 and 36) is:

std::string
toString( unsigned n, int precision, unsigned base )
{
    assert( base >= 2 && base <= 36 );
    static char const digits[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    std::string retval;
    while ( n != 0 ) {
        retval += digits[ n % base ];
        n /= base;
    }
    while ( retval.size() < precision ) {
        retval += ' ';
    }
    std::reverse( retval.begin(), retval.end() );
    return retval;
}

You can then display it.

score 0 · Answer 6 · answered May 21 '12 at 12:43

Recursion. In pseudocode:

function toBinary(integer num)
  if (num < 2) 
  then
    print(num)
  else
    toBinary(num DIV 2)
    print(num MOD 2)
  endif
endfunction

This does not handle leading zeros or negative numbers. The recursion stack is used to reverse the binary bits into the standard order.

score -1 · Answer 7 · edited Aug 11 '13 at 16:01

-1

Just write:

long int dec,rem,i=1,sum=0

Instead of:

int dec,rem,i=1,sum=0;

That should solve the problem.

edited Aug 11 '13 at 16:01

Andre Silva

4,782
9
52
65

answered Aug 11 '13 at 15:20

Milind

1

1

hmm ... _long int_ - are you sure? Anyway, please edit your answer and format the code (4 spaces at line start, ctrl-k is your friend :) – kleopatra Aug 11 '13 at 15:31

convert decimal to 32 bit binary?

7 Answers7