Efficient way to print int in binary format

void printBinary(int *intVal){
  register unsigned int mask;
  mask = 1<<((8*sizeof(int))-1);
    print("%d",((*intVal & mask)/mask));
  while((mask >>= 1) & -1);

Ok, Sorry I did not put any comments in the code, simply because I want it to look short and sweet. Now let me elaborate a bit on how this works.

void printBinary(int *intVal){

We are taking a pointer here, so make sure your calling function sends an address and not the value.

  register unsigned int mask;

Register for speed, we need to have an unsigned int.

mask = 1<<((8*sizeof(int))-1);

Ok, let me explain this one step at a time
Step 1: Get Size of integer.
Step 2: Multiply it with 8 ( we need the Size in Bits, not in bytes).
Step 3: Subtract this value by 1
Step 4 : Left Shift 1 by the value obtained by Step 3 calculation.

How does this work?
Assume that the platform where you are gonna run this code represents binary as 4bytes. So
Step 1: 4 Bytes
Step 2 : 4×2 = 32 Bits
Step 3 : 32 -1 = 31
Step 4 : Left shift 1 (binary : 00000000000000000000000000000001) by 31 times = binary : 10000000000000000000000000000000

    print("%d",((*intVal & mask)/mask));
  while((mask >>= 1) & -1);

Now comes the bit tricky part, so how do we do this? Quite simple once you get a hang of it.

First we do a bitwise and on the value we got from the calling function and the mask. We then divide it by mask.

Initially, the mask will be at the leftmost position (which is usually the sign bit, hence will be 1 if the value is negative). Now if the number you passed on was signed and negative, then ‘and’ing with mask will ensure that only signed bit will remain and makes the rest of the bits as ‘0’. Now you divide this value by mask itself. Hence you will get value with 1.

Now if the bit was zero then the result of  ‘and’ing would have been all zeros. Thus dividing the value with mask would have yielded 0.

Ok, now the next step, we right shift the mask by 1. So we will end up with 01000000000000000000000000000000 , first time around. We then do a bitwise and with -1 which is represented as binary all 1’s. Thus this while loop will remain true till we keep right shifting 1 in the mask all the way to the right, and finally dropped. By that time we would have printed the value of  referenced by intVal pointer in binary form.

Simple? 🙂

Tip: If you know the size of integer in your platform then you can get some more performance and simplicity by pre calculating what number will be 1 followed by all zeros. For 4 bytes this value will be 2147483648.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s