Arredonda o valor decimal no awk

Question

eu quero arredondar o valor 262,5 para 263 usando o awk

eu tentei o abaixo

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

Minha resposta esperada é 263

    

Answer 1

De acordo com o manual do GNU awk ,

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.

A mesma página também tem um exemplo 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 2

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