Round off the decimal value in awk

Question

i want to round off the value 262.5 to 263 using awk

i have tried the below

echo "262.5" | awk '{printf "%.f\n", $1}'

My expected answer is 263

Answer 1

Little different for positive numbers:

echo "262.5" | awk '{ print int($1  + 0.5) } '

(Result: 263)

echo "262.4" | awk '{ print int($1  + 0.5) } '

(Result: 262)

Rounding on 1st decimal place:

echo " " | awk '{ print int(5.89 * 0.3851 * 10 + 0.5)/10 } '

(Result: 2.3)

Rounding on 2nd decimal place:

echo " " | awk '{ print int(5.21 * 0.3851 * 100 + 0.5)/100 } '

(Result: 2.01)

And so on...

Answer 2

According to the GNU awk manual,

The way printf and sprintf() (see Printf) perform rounding often depends upon the system's C sprintf() subroutine. On many machines, sprintf() rounding is “unbiased,” which means it doesn't always round a trailing ‘.5’ up, contrary to naive expectations. In unbiased rounding, ‘.5’ rounds to even, rather than always up, so 1.5 rounds to 2 but 4.5 rounds to 4.

The same page also has an example round() function:

function round(x, ival, aval, fraction)
{
   ival = int(x)    # integer part, int() truncates

   # see if fractional part
   if (ival == x)   # no fraction
      return ival   # ensure no decimals

   if (x < 0) {
      aval = -x     # absolute value
      ival = int(aval)
      fraction = aval - ival
      if (fraction >= .5)
         return int(x) - 1   # -2.5 --> -3
      else
         return int(x)       # -2.3 --> -2
   } else {
      fraction = x - ival
      if (fraction >= .5)
         return ival + 1
      else
         return ival
   }
}

Answer 3

echo "262.5" | awk '{printf "%.f\n", int($1+0.5)}'