Bash printing diagonal of any size array

Question

So here I have the following code:

#!/bin/bash
awk \'BEGIN {f=4} {printf($f\" \"); f=f-1}\'

Which will take input such as:

Answer 1

awk '!i{i=NF}{printf "%s ",$i;i--}'

EDIT: Since Ed has already exploited all the nice and proper awk solutions I'll throw in a native bash solution:

$ cat t.sh
#!/bin/bash

while read -ra numbers; do
        : ${i:=${#numbers[@]}}
        diag+=("${numbers[--i]}")
done < test.txt
echo "${diag[@]}"

.

$ ./t.sh
4 7 1 4

Answer 2

The problem is, NF at BEGIN section is not the number of tokens on the first line. The following works:

awk '{if (!f) f = NF; printf($f" "); f=f-1}'

Edit: as per suggestions in the comments, a cleaner and safer way is:

awk '{if (!f) f = NF; printf("%s ", $f); f--}'

Answer 3

$ awk '{print $(NF+1-NR)}' file    
4
7
1
4

$ awk -v ORS=" " '{print $(NF+1-NR)}' file
4 7 1 4 

or if you want to avoid adding a space to the end of your output line and to have a terminating newline:

$ awk '{printf "%s%s", (NR>1?FS:""), $(NF+1-NR)} END{print ""}' file
4 7 1 4