A Shell Script to simulate the wc command with its options? - linux

Question

we have to write a shell script program , which works similar to wc command.
receives -l, -c and -w as its options.
Shell scripting syntax aside; MY QUESTION is that can we simulate logic of wc -c or wc -l or wc -w using sed or grep or anything else ; if yes then how?
IMP: Don't use wc in script

Answer 1

A single awk command that you can parameterize by setting the appropriate -v variables to 0:
LC_ALL=C awk -v l=1 -v w=1 -v c=1 '
{ wc+=NF; cc+=1+length($0) }
END { printf "%s\t%s\t%s\n", l ? NR : "", w ? wc: "", c ? cc : ""}
' file
Note:
For simplicity, you always get 3 \t-separated output fields, with fields whose output wasn't requested empty; it wouldn't be hard to modify this to emulate wc's output behavior, however.
As explained in choroba's grep answer, you must prepend LC_ALL=C  to awk ..., if you really want to count bytes (-c) rather than (potentially multi-byte) characters (-m).
To count characters (the equivalent of wc -m), remove LC_ALL=C  above.
Caveat: This won't work BSD awk, as also found on macOS, unfortunately, because it not Unicode-aware and always counts the number of bytes (try awk '{print length($0)}' <<<ü).
wc -l strictly counts the number of \n characters, so it doesn't count an incomplete line - one missing a trailing \n - at the end of its input; the above awk command, by contrast, does count that line (and an implied trailing newline in the byte/character count).
How it works:
awk's NF variables contains the number of fields on each input line, where the line is broken into fields by arbitrary runs of whitespace by default; in other words: by default, fields are words.
$0 is the input line at hand, whose length() tells you the number of characters / bytes, with 1 added to account for the \n character at the end of the line.
Note how variables wc and cc need to initialization, because awk implicitly treats empty/undefined variables as 0 in a numeric context (such as with compound operator +=).
NR contains the current, 1-based line number, which in the END block is equal to the total number of input lines.

Answer 2

Using awk:
-l:
awk 'END{print NR}' inFile
-w:
awk '{words+=NF}END{print words}' inFile
-c:
ls -l inFile | awk '{print $5}'

Answer 3

If you can use grep, simulating the line count is easy: just count how many times something that matches always happens:
grep -c '^' filename
This should output the same as wc -l (but it might report one more line if the file doesn't end in a newline).
To get the number of words, you can use the following pipeline:
grep -o '[^[:space:]]\+' filename | grep -c '^'
You need grep that supports the -o option which prints each matching string to a line of its own. The expression matches all non-space sequences, and piping them into what we used in the previous case just counts them.
To get the number of characters (wc -c), you can use
LC_ALL=C grep -o . filename | grep -c '^'
Setting LC_ALL is needed if your locale supports UTF-8, otherwise you'd count wc -m. You need to add the number of newlines to the output number, so
echo $(( $( grep -c '^' filename )
+ $( LC_ALL=C grep -o . filename | grep -c '^' ) ))