Awk if-statement to count the number of characters (wc -m) coming from a pipe-CodePudding

I tried to scratch my head around this issue and couldn't understand what it wrong about my one liner below.

Given that

echo "5" | wc -m 
2

and that

echo "55" | wc -m
3

I tried to add a zero in front of all numbers below 9 with an awk if-statement as follow:

echo "5" |  awk '{ if ( wc -m $0 -eq 2 ) print 0$1 ; else print $1 }'
05

which is "correct", however with 2 digits numbers I get the same zero in front.

echo "55" |  awk '{ if ( wc -m $0 -eq 2 ) print 0$1 ; else print $1 }'
055

How come? I assumed this was going to return only 55 instead of 055. I now understand I'm constructing the if-statement wrong.

What is then the right way (if it ever exists one) to ask awk to evaluate if whatever comes from the | has 2 characters as one would do with wc -m?

I'm not interested in the optimal way to add leading zeros in the command line (there are enough duplicates of that).

Thanks!

CodePudding user response：

I suggest to use printf:

printf "d\n" "$(echo 55 | wc -m)"
03
printf "d\n" "$(echo 123456789 | wc -m)"
10

Note: printf is available as a bash builtin. It mainly follows the conventions from the C function printf().. Check

help printf   # For the bash builtin in particular
man 3 printf  # For the C function

CodePudding user response：

Facts:

In AWK strings or variables are concatenated just by placing them side by side.
For example: awk '{b="v" ; print "a" b}'
In AWK undefined variables are equal to an empty string or 0.
For example: awk '{print a "b", -a}'
In AWK non-zero strings are true inside if.
For example: awk '{ if ("a") print 1 }'

wc -m $0 -eq 2 is parsed as (i.e. - has more precedence then string concatenation):

wc -m $0 -eq 2
( wc - m ) ( $0 - eq ) 2
                       ^   - integer value 2, converted to string "2"
                  ^^       - undefined variable `eq`, converted to integer 0
             ^^            - input line, so string "5" converted to integer 5
                ^          - subtracts 5 - 0 = 5
            ^^^^^^^^^^^    - integer 5, converted to string "5"
       ^                   - undefined variable "m", converted to integer 0
  ^^                       - undefined variable "wc" converted to integer 0
^^^^^^^^^                  - subtracts 0 - 0 = 0, converted to a string "0"
^^^^^^^^^^^^^^^^^^^^^^^^^  - string concatenation, results in string "052"

The result of wc -m $0 -eq 2 is string 052 (see awk '{ print wc -m $0 -eq 2 }' <<<'5'). Because the string is not empty, if is always true.

It should return only 55 instead of 055

No, it should not.

Am I constructing the if statement wrong?

No, the if statement has valid AWK syntax. Your expectations to how it works do not match how it really works.

CodePudding user response：

To actually make it work (not that you would want to):

echo 5 | awk '
{
  cmd = "echo " $1 " | wc -m"
  cmd | getline len
  if (len == 2)
    print "0"$1
  else
    print $1
}'

But why when you can use this instead:

echo 5 | awk 'length($1) == 1 { $1 = "0"$1 } 1'

Or even simpler with the various printf solutions seen in the other answers.