unix - How to print ASCII value of a character using basic awk only

Question

I need to print the ASCII value of the given character in awk only.

Below code gives 0 as output:

echo a | awk '{ printf("%d 
",$1); }'

Answer 1

Using only basic awk (not even gawk, so the below should work on all BSD and Linux variants):

$ echo a | awk 'BEGIN{for(n=0;n<256;n++)ord[sprintf("%c",n)]=n}{print ord[$1]}'
97

Here's the opposite direction (for completeness):

$ echo 97 | awk 'BEGIN{for(n=0;n<256;n++)chr[n]=sprintf("%c",n)}{print chr[$1]}'
a

Basic premise is to use a lookup table.