How to get printf to treat the variables passed from bc script differently?

Question

I'm writing a program for CodeWars that requires me to return 2 specific variables with a space between them.

The following is the script that I am trying to execute

!/bin/bash

nbMonths()
{
    printf "%.0f\n" $(
    bc <<< "
      scale=5
      osum=$1
      nsum=$2
      sav=0
      lper=($4 / 100)
      cmon=0

      diff = osum + sav - nsum

      while( diff < 0 ) {
         osum -= ( osum * lper )
         nsum -= ( nsum * lper )
         sav += $3
         cmon += 1
      
         scale=0
         if( (cmon%2) == 1) {
           scale=5
           lper += 0.005
         }
    
         scale=5
         diff = osum + sav - nsum
      } 

      print cmon, diff
    ")
}

nbMonths $1 $2 $3 $4

For the following input to the bash script

bash nameOfProg.sh 2000 8000 1000 1.5

The result it shows is

6766

While I would like it to display it as

6 766

There are 2 questions I have regarding this,

1. Is there a way to make printf to print in the above format?

2. I've noticed that print cmon, \", \", diff results in

   6, 766

   Are there are escape sequences(like \",\") that I could use? If so, where could I find it since there isn't a man page for print.

Please note that the reason I've used %0.f in the printf statement is that I need the last value (from diff) to be rounded off.

Answer 1

you can print a space in a bc print command by enquoting the space in double strings.

Since your bc input is already enquoted, you'll need to escape the quotes with backslashes so it doesn't end the here-string (data following <<<).

Here's a simple example:

$  bc <<<"print 1, \" \", 3, \" \", 4"; echo
1 3 4

Answer 2

While I would like it to display it as

Then:

1. Remove printf

2. Read bc documentation and note that strings start and end with double quotes.

3. Learn about shell quoting. If you are inside " quotes, then to pass " you have to escape it.

   $ cat <<<"inside double quotes: literal quote: \" "
   inside double quotes: literal quote: "
   

4. Print the space between numbers.

#!/bin/bash
bc <<<"
  scale=5
  osum=$1
  nsum=$2
  sav=0
  lper=($4 / 100)
  cmon=0

  diff = osum + sav - nsum

  while( diff < 0 ) {
     osum -= ( osum * lper )
     nsum -= ( nsum * lper )
     sav += $3
     cmon += 1
  
     scale=0
     if( (cmon%2) == 1) {
       scale=5
       lper += 0.005
     }

     scale=5
     diff = osum + sav - nsum
  } 

   print cmon, \" \", diff
"