How to round up on fives in unix?


 
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# 1  
Old 11-18-2009
Error How to round up on fives in unix?

i'm a newbie here, i need help with a shell script.

for a given number, if it is greater than ten round to the nearest 10
same for 100, if it is greater than 100 round to the nearest 100, and same for 1000.
i'm confused how to start this...

its supposed to look like this

input ----- output
15 ----- 20
99 ----- 100
12345678 ----- 10000000
44444445 ---- 50000000
1445 ---- 2000

i seem to get lost with this.
any help how to do this
# 2  
Old 11-18-2009
Hi, as an alternative to calculus you could use string operations and a simple addition:
  • Take the first digit and add 1.
  • Replace the rest of the digits with zeroes
  • Stick them together.
Here is an example of how this can be achieved. If you find it a little overwhelming, just try your own code. There are many ways to get from a to b.

Code:
while read i ;do
   x=$((i-${i#?}))
   echo $((${x%%0*}+1))${x#?}
done < infile


Last edited by Scrutinizer; 11-18-2009 at 05:32 PM..
# 3  
Old 11-19-2009

Code:
printf "%s\n" 15 99 12345678 44444445 1445 |
awk '{
  n = sprintf( "%1.0e\n", $1)
  printf "%15d -- %d\n", $1, n
}'

# 4  
Old 11-19-2009
Code:
awk '
{
  x = 10
  do {
    $0 = int(($0 + x/2)/x)*x
  } while ((x = x*10) < $0)
  print
}'

# 5  
Old 11-19-2009
Code:
#!/bin/ksh93

for x in 15 99 12345678 44444445 1445
do
    printf "%15d -- %d\n" $x $(( rint( $(printf "%1.0e" $x) ) ))
done

# 6  
Old 11-20-2009
ok

thank you, but i entered this in to a script file and i doesnt do anything.
and the input numbers are sample one, im supposed to read numbrs from a command.
like this
./script 23
then it rounds in off, they are not set numbers.
# 7  
Old 11-20-2009
Code:
printf "%s\n" "$@" |
awk '{
  n = sprintf( "%1.0e\n", $1)
  printf "%d\n", n
}'

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