Tricky for loop

[hzcodec]

I found a piece of code a while ago with a for loop that I don't understand. The purpose is counting the number of bits.

#include <stdio.h>

int main()
{
  unsigned int i;
  unsigned int v = 14;

  for (i=0; v; v>>=1) {
    i+= v & 1;
  }

  printf("# bits %d\n",i);

  return 0;
}

The question I have is what is happening here?
Especially the code in the "for" statement, (some shifting):

v>>=1

and later on (ANDing one bit or whole vector?):

i+= v & 1;

I have no problem understanding bits and bytes. The result here is of course 3 bits.

[Salem]

Perhaps print the value in v each time around the loop (in binary for extra points), and also print it when the loop exits.

[hzcodec]

I have of course done this. However I still have problem understanding the "v>>=1".

[laserlight]

However I still have problem understanding the "v>>=1".

Read up on bit shifting.

[hzcodec]

Well ... "v>>1" is shifting one step but what is the "=" sign doing in the for loop?

[laserlight]

Well ... "v>>1" is shifting one step but what is the "=" sign doing in the for loop?

You may be familiar with syntax like:

x += 2;

as shorthand for:

x = x + 2;

The same idea applies here.

[hzcodec]

Oh yes! I'm familiar with the shorthands. And with a little a bit of experimenting with the code I finally got it. Just confusing myself. laserlight got me thinking in the right direction. Thank you for your help.