how does this work in awk even without print?

Question

bash-3.00# cat > b1.txt
AAA,Apples,123
BBB,Bananas,124
CCC,Carrot,125
bash-3.00# cat > b2.txt
Store1|AAA|123|11
Store2|BBB|124|23
Store3|CCC|125|57
Store4|DDD|126|38
bash-3.00# awk -F"[,|]" 'NR==FNR{a[$1]++;next} !a[$2]' b1.txt b2.txt
Store4|DDD|126|38

When NR==FNR returns true for first file a[$1] gets incremented. next brings back to beginning of the script. But what is !a[$2] doing here? How does it print without even a print command?

Answer 1

From the POSIX awk spec:

An awk program is composed of pairs of the form:

pattern { action }

Either the pattern or the action (including the enclosing brace characters) can be omitted. ... [A] missing action shall be equivalent to:

{ print }

In your case, !a[$2] is the pattern, and the action is missing.

Answer 2

To expand on the previous answer:

awk handles true/false tests numerically; 0 is "false" and any other number is "true". The exclam (!) reverses the test.

So !a[$1] is essentially equivalent to saying, "print if a[$1] is equal to 0".