How to iterate through all ASCII characters in Bash? - linux

Question

I know how to iterate through alphabets:
for c in {a..z}; do ...; done
But I can't figure out how to iterate through all ASCII characters. Does anyone know how?

Answer 1

What you can do is to iterate from 0 to 127 and then convert the decimal value to its ASCII value(or back).
You can use these functions to do it:
# POSIX
# chr() - converts decimal value to its ASCII character representation
# ord() - converts ASCII character to its decimal value
chr() {
[ ${1} -lt 256 ] || return 1
printf \\$(printf '%03o' $1)
}
# Another version doing the octal conversion with arithmetic
# faster as it avoids a subshell
chr () {
[ ${1} -lt 256 ] || return 1
printf \\$(($1/64*100+$1%64/8*10+$1%8))
}
# Another version using a temporary variable to avoid subshell.
# This one requires bash 3.1.
chr() {
local tmp
[ ${1} -lt 256 ] || return 1
printf -v tmp '%03o' "$1"
printf \\"$tmp"
}
ord() {
LC_CTYPE=C printf '%d' "'$1"
}
# hex() - converts ASCII character to a hexadecimal value
# unhex() - converts a hexadecimal value to an ASCII character
hex() {
LC_CTYPE=C printf '%x' "'$1"
}
unhex() {
printf \\x"$1"
}
# examples:
chr $(ord A) # -> A
ord $(chr 65) # -> 65

Answer 2

A possibility using only echos octal escape sequences:
for n in {0..7}{0..7}{0..7}; do echo -ne "\\0$n"; done

Answer 3

Here is what I came up with for a one-liner taking some pieces from sampson-chen's and mata's answers:
for n in {0..127}; do awk '{ printf("%c", $0); }' <<< $n; done
Or alternatively:
for n in {0..127}; do echo $n; done | awk '{ printf("%c", $0); }'

Answer 4

Here's how you print an integer as its corresponding ascii character with awk:
echo "65" | awk '{ printf("%c", $0); }'
which will print:
A
And here's how you might iterate through the uppercase alphabet this way:
# ascii for A starts at 65:
ascii=65
index=1
total=26
while [[ $total -ge $index ]]
do
letter=$(echo "$ascii" | awk '{ printf("%c", $0); }')
echo "The $index'th letter is $letter"
# Increment the index counter as well as the ascii counter
index=$((index+1))
ascii=$((ascii+1))
done

Answer 5

Well... If you really want them all, and you want it to be something script-like, you could do this, I guess:
awk 'function utf32(i) {printf("%c%c%c%c",i%0x100,i/0x100%0x100,i/0x10000%0x100,i/0x1000000) } BEGIN{for(i=0;i<0x110000;i++){utf32(i);utf32(0xa)}}' | iconv --from-code=utf32 --to-code=utf8 | grep -a '[[:print:]]'
But the list is pretty huge, and not very useful. Awk may not be the most elegant way of generating the binary integers from 0 to 0x110000 - substitute something more elegant if you find it.
Edit: Oh, I see you only wanted ascii. Well, I will let this answer stay here in case somebody else actually wants all the UTF printable characters.

Answer 6

It depends what you mean by iterate. Be aware that NUL can't be assigned or passed to commands.
This generates all ascii characters
seq 0 127 |\
xargs printf '\\x%x ' |\
xargs printf '%b '
seq 0 127 generate all integers between 0 and 127
xargs printf '\\x%x ' convert it to hex, separated by space
xargs printf '%b ' convert hex to byte, separated by space