awk - What does a number do after curly braces?

Question

Why does

echo foo bar..baz bork | awk 'BEGIN{RS=".."} {gsub(OFS,"");}1'

seem to do the same thing as

echo foo bar..baz bork | awk 'BEGIN{RS=".."} {gsub(OFS,"");} {print;}'

?

In fact any number that isn't zero (including decimals and negatives) will do the same thing. However, leaving off the digit, using a text character, or using zero prints nothing. I didn't see this documented anywhere, although I could have missed something.

Answer 1

If you remember, awk is a language which has a series of <pattern> <action> operations. Each pattern is evaluated for each line (at least conceptually), and when the pattern matches, the action is executed. Either the pattern or the action can be omitted. An omitted pattern matches every line; an omitted action defaults to {print $0} (aka {print}). The 'pattern' might be a simple regex match, or some other more complicated and general condition, which must evaluate to true if the action is to be executed (as noted by Ed Morton in his comment).

In your example, the 1 is a pattern; it evaluates to true. The action is not specified, so the default action is invoked, which is {print} or {print $0}. Any value other than zero or an empty string evaluates to true and will invoke the print. (Note that if you mention an uninitialized variable (for example, c), then it is autocreated and set to zero and therefore evaluates to false. Hence awk 'c' <<<"Hi" prints nothing.)

The actions associated with the BEGIN and END patterns are handled specially, of course.