Why does awk "not in" array work just like awk "in" array?

Question

Here's an awk script that attempts to set difference of two files based on their first column:

BEGIN{
    OFS=FS="\t"
    file = ARGV[1]
    while (getline < file)
        Contained[$1] = $1
    delete ARGV[1]
    }
$1 not in Contained{
    print $0
}

Here is TestFileA:

cat
dog
frog

Here is TestFileB:

ee
cat
dog
frog

However, when I run the following command:

gawk -f Diff.awk TestFileA TestFileB

I get the output just as if the script had contained "in":

cat
dog
frog

While I am uncertain about whether "not in" is correct syntax for my intent, I'm very curious about why it behaves exactly the same way as when I wrote "in".

Answer 1

I cannot find any doc about element not in array.

Try !(element in array).

---

I guess: awk sees not as an uninitialized variable, so not is evaluated as an empty string.

$1 not == $1 "" == $1 

Answer 2

I figured this one out. The ( x in array ) returns a value, so to do "not in array", you have to do this:

if ( x in array == 0 )
   print "x is not in the array"

or in your example:

($1 in Contained == 0){
   print $0
}

Answer 3

In my solution for this problem I use the following if-else statement:

if($1 in contained);else{print "Here goes your code for \"not in\""}

Answer 4

Not sure if this is anything like you were trying to do.

#! /bin/awk
# will read in the  second arg file and make a hash of the token
# found in column one. Then it will read the first arg file and print any
# lines with a token in column one not matching the tokens already defined
BEGIN{
    OFS=FS="\t"
    file = ARGV[1]
    while (getline  &lt file)
        Contained[$1] = $1
#    delete ARGV[1]  # I don't know what you were thinking here
#    for(i in Contained) {print Contained[i]} # debuging, not just for sadists
    close (ARGV[1])
}
{
   if ($1 in  Contained){} else { print $1 }
}