Question

I can’t understand how the following code works.

$start = 1;
while($start<10){
    if ($start&1) {
      echo "ODD  ".$start." <br/> ";
    }
    else {
      echo "EVEN  ".$start." <br/> ";
    }
    $start++;
}

The $start&1 will return ODD and EVEN seperately.

Output

ODD 1
EVEN 2
ODD 3
EVEN 4
ODD 5
EVEN 6
ODD 7
EVEN 8
ODD 9 

If we give $start&2 instead of $start&1, it returns with another order.

How &1 &2 etc… works here?

Answer 1

It is a bitwise and operator.

  0001 --> 1
& 0001
  ----
  0001 --> 1

  0010 --> 2
& 0001
  ----
  0000 --> 0

  0011 --> 3
& 0001
  ----
  0001 --> 1

Depending on the endianness, it will be either the leftmost or rightmost bit that matters in this check. The above is &ing with 1. In your second example, &ing with 2, that would be

  0001 --> 1
& 0010
  ----
  0000 --> 0

  0010 --> 2
& 0010
  ----
  0010 --> 2

  0011 --> 3
& 0010
  ----
  0010 --> 2

And for further comparison, here is 1-3 &ing with 3

  0001 --> 1
& 0011
  ----
  0001 --> 1

  0010 --> 2
& 0011
  ----
  0010 --> 2

  0011 --> 3
& 0011
  ----
  0011 --> 3

To see what is going on, follow the columns of the two numbers down. If they are both a 1 then the result has the bit set in that position to a 1. If either are a 0 then the result is a 0 in that position. So for 2 & 3..

  0010 --> 2
& 0011
  ----
  0010
  ||||
  |||+- 0 and 1, so 0
  ||+-- 1 and 1, so 1
  ++--- 0 and 0, so 0

  0010 == 2