how to loop through the digits of a binary number?

Question

I have a binary number 1011011, how can I loop through all these binary digits one after the other ?

I know how to do this for decimal integers by using modulo and division.

Answer 1

int n = 0x5b; // 1011011

Really you should just do this, hexadecimal in general is much better representation:

printf("%x", n); // this prints "5b"

To get it in binary, (with emphasis on easy understanding) try something like this:

printf("%s", "0b"); // common prefix to denote that binary follows
bool leading = true; // we're processing leading zeroes
// starting with the most significant bit to the least
for (int i = sizeof(n) * CHAR_BIT - 1; i >= 0; --i) {
    int bit = (n >> i) & 1;
    leading |= bit; // if the bit is 1, we are no longer reading leading zeroes
    if (!leading)
        printf("%d", bit);
}
if (leading) // all zero, so just print 0
    printf("0");

// at this point, for n = 0x5b, we'll have printed 0b1011011

Answer 2

You can use modulo and division by 2 exactly like you would in base 10. You can also use binary operators, but if you already know how to do that in base 10, it would be easier if you just used division and modulo

Answer 3

Expanding on Frédéric and Gabi's answers, all you need to do is realise that the rules in base 2 are no different to in base 10 - you just need to do your division and modulus with a divisor 2 instead of 10.

The next step is simply to use number >> 1 instead of number / 2 and number & 0x1 instead of number % 2 to improve performance. Mind you, with modern optimising compilers there's probably no difference...

Answer 4

Use an AND with increasing powers of two...

Answer 5

In C, at least, you can do something like:

while (val != 0)
{
   printf("%d", val&0x1);
   val = val>>1;
}

Answer 6

To expand on @Marco's answer with an example:

uint value = 0x82fa9281;

for (int i = 0; i < 32; i++)
{
    bool set = (value & 0x1) != 0;
    value >>= 1;

    Console.WriteLine("Bit set: {0}", set);
}

What this does is test the last bit, and then shift everything one bit.

Answer 7

If you're already starting with a string, you could just iterate through each of the characters in the string:

var values = "1011011".Reverse().ToCharArray();
for(var index = 0; index < values.Length; index++) {
 var isSet = (Boolean)Int32.Parse(values[index]); // Boolean.Parse only works on "true"/"false", not 0/1
 // do whatever
}

Answer 8

        byte input = Convert.ToByte("1011011", 2);
        BitArray arr = new BitArray(new[] { input });
        foreach (bool value in arr)
        {
            // ...
        }

Answer 9

You can simply loop through every bit. The following C like pseudocode allows you to set the bit number you want to check. (You might also want to google endianness)

for()
{
  bitnumber = <your bit>
  printf("%d",(val & 1<<bitnumber)?1:0);
}

The code basically writes 1 if the bit it set or 0 if not. We shift the value 1 (which in binary is 1 ;) ) the number of bits set in bitnumber and then we AND it with the value in val to see if it matches up. Simple as that!

So if bitnumber is 3 we simply do this

00000100 ( The value 1 is shifted 3 left for example)

AND

10110110 (We check it with whatever you're value is)

=

00000100 = True! - Both values have bit 3 set!