Binary Representation

**clag** · 01-19-2010

Code:

void printOnes (int x) 
{ 
  int i; 
  int numOnes = 0; 
  for (0 = i; i < 32; i++) { 
    if ((x & 0x1) = 1) 
      numOnes++; 
    x = x >> 1; 
  } 
  printf("The number of ones is %c\n", numOnes); 
}

This is supposed to count the number of 1s in the binary representation of an integer. I know there are some logical errors.

I know the logical operator x & 0x1 simply seeing if it's odd but I don't see how this counts the number of 1s. (isn't is just saying x % 2 == 1. so x could be 3,5..etc) How would I fix this?

or is x=x>>1 wrong?

**iMalc** · 01-19-2010

= 1 should be == 1

**hk_mp5kpdw** · 01-19-2010

Code:

  for (0 = i; i < 32; i++) {

Does that even compile?

**EVOEx** · 01-19-2010

Originally Posted by hk_mp5kpdw

Code:

  for (0 = i; i < 32; i++) {

Does that even compile?

Wait, what's wrong with changing The World so that 0 gets another value? ;-)

Really, the two problems have been pointed out. I just wanted to note that there is another algorithm that can be faster:

Code:

count = 0
while number != 0
  number = number & (number - 1)
  count++

Actually, it's a well known algorithm. But I don't know if it's actually faster. It requires less steps (or as many steps) to find the answer, but decrementing might require more time than decrementing...
Anyways, it's just informative, it's a fun algorithm. See if you can figure out how it works.

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