CODEWITHSUNDEEP
awk
William Pursell
William Pursell

Reputation: 212464

post-decrement of NF in awk

I'm a bit confused by the following:

$ echo foo bar baz | awk '{printf "%d:", NF--; print NF}'
3:2
$ echo foo bar baz | awk '{printf "%d:", NF; $NF=""; NF -= 1; print NF}' 
3:2
$ echo foo bar baz | awk '{printf "%d:", NF; $(NF--)=""; print NF}' 
3:3

I see the same behavior in awk version 20070501 (macos) and GNU Awk 4.0.2. Why does the post-decrement of NF in the 3rd case not apply? Is that behavior expected, mandated by a standard, or a quirk of the implementation?

EDIT by Ed Morton: FWIW I'd find the following a more compelling example:

$ echo foo bar baz | awk '{printf "%d:", NF; NF--; $NF=""; print NF}'
3:2

$ echo foo bar baz | awk '{printf "%d:", NF; --NF; $NF=""; print NF}'
3:2

$ echo foo bar baz | awk '{printf "%d:", NF; $NF=""; NF--; print NF}'
3:2

$ echo foo bar baz | awk '{printf "%d:", NF; $NF=""; --NF; print NF}'
3:2

$ echo foo bar baz | awk '{printf "%d:", NF; $(--NF)=""; print NF}'
3:2

$ echo foo bar baz | awk '{printf "%d:", NF; $(NF--)=""; print NF}'
3:3

with the question being why does the last example (post-decrement with assignment) behave differently from all of the other cases, regardless of which one you think it should be equivalent to.

Upvotes: 4

Views: 173

Answers (1)

Barmar
Barmar

Reputation: 781848

The value of a post-decrement is the value of the variable before it's decremented. Because of this, the last case is adding a new field after it decrements NF, which updates NF.

$(NF--) = "";

is equivalent to

temp = NF;  # temp == 3
NF--;       # NF == 2
$temp = ""; # adds a new field 3, so now NF == 3

Upvotes: 7

Related Questions