Shell: why does while loop processing additional char

Question

This is my simple shell script

root@Ubuntu:/tmp# cat -n script.sh 
     1  echo
     2  while x= read -n 1 char
     3  do
     4   echo -e "Original value = $char"
     5   echo -e "Plus one = `expr $char + 1`\n"
     6  done < number.txt
     7  echo
root@Ubuntu:/tmp#

And this is the content of number.txt

root@Ubuntu:/tmp# cat number.txt 
12345
root@Ubuntu:/tmp#

As you can see on the code, I'm trying to read each number and process it separately. In this case, I would like to add one to each of them and print it on a new line.

root@Ubuntu:/tmp# ./script.sh 

Original value = 1
Plus one = 2

Original value = 2
Plus one = 3

Original value = 3
Plus one = 4

Original value = 4
Plus one = 5

Original value = 5
Plus one = 6

Original value = 
Plus one = 1


root@Ubuntu:/tmp#

Everything looks fine except for the last line. I've only have 5 numbers, however it seems like the code is processing additional one.

Original value = 
Plus one = 1

Question is how does this happen and how to fix it?

halfbit · Accepted Answer · 2017-11-01 09:04:45Z

1

It seems the input file number.txt contains a complete line, which is terminated by a line feed character (LF). (You can verify the input file is longer than 5 using ls -l.) read eventually encounters the LF and gives you an empty char (stripping the terminating LF from the input as it would without the -n option). This will give you expr + 1 resulting in 1. You can explicitely test for the empty char and terminate the while loop using the test -n for non-zero length strings:

echo "12345" | while read -n 1 char && [ -n "$char" ]; do echo "$char" ; done

answered Nov 1, 2017 at 9:04

halfbit

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