Shell Programming and Scripting

I am trying to write the following c code in nawk:

while ((ch = getc (fp)) != EOF)
{
total_bytes++;
checksum = (checksum >> 1) + ((checksum & 1) << 15);
checksum += ch;
checksum &= 0xffff; /* Keep it within bounds. */
}

I appreciate your help

My nawk man pages do not have and "&=" or ">> <<" shift left or shift right operators.
nawk apparently does not support that. You can fake >> and << with division by 2 and multiplication by two. I don't know a way to fake the "and" operation without writing
some very tricky code. You are doing 32 bit operations. That means "anding" 32 values.

Try cksum or write your own C code. That makes more sense.

Maybe one of our ultimo awkers have done this solidly, but it looks like a nightmare waiting to happen to me.

Thanks for your prompt response. We will see if someone can help

Thanks again

yea my problem is how to do the AND.

Here is your problem, why no "&" operator:

really old awk implementations used single precision floating point for numbers.
Newer awks, like nawk, use double precision. None use 32 bit integers.

the most recent 'gawk' supports bitwise ops: and, or, xor, compl, lshift and rshift
here's something found seaching comp.lang.awk:

Code:

function awkand(a,b,  k,r) {
  k = 1
  r = 0
  a = int(a + 0.5) # round to nearest integer
  b = int(b + 0.5)
  while ( a + b ) {
    if ( a%2 && b%2 ) r += k
    a = int(a / 2)
    b = int(b / 2)
    k *= 2
  }
  return r

}

And'ing a number with FFFF is the same as finding the remainder when dividing by 65535. This operation is called modulo and could be implemented as:
$ echo "65536
> 65537
> 65538" | awk 'function mod(v,m) { return v-(int(v/m)*m) }
> { print $1, mod($1,65535) }'
65536 1
65537 2
65538 3
$