How to Print an ASCII Character by Different Code Points in Bash

There's a printf tool that simulates the C function; normally it's at /usr/bin/printf, but a lot of shells implement built-ins for it as well. You can use %02x to get the hex representation of a character, but you need to make sure you pass a string that includes the character in single-quotes (Edit: It turns out just a single-quote at the beginning is sufficient):

printf "%%%02x\n" "'-"   # Outputs %2d

You can make a shell function for convenience:

function hex() {
    printf "%%%02x\n" "'$1"
}

hex -   # Outputs %2d
hex _   # Outputs %5f
hex .   # Outputs %2e

You can't directly print the ascii codes by using the printf "%c" $i like in C.

You have to first convert the decimal value of i into its octal value and then you have to print it using using printf and putting \ in front of their respective octal values.

To print A, you have to convert the decimal 65 into octal, i.e. 101, and then you have to print that octal value as:

printf "\101\n"

This will print A.

So you have to modify it to :

for i in `seq 32 127`; do printf \\$(printf "%o" $i);done;

But by using awk you can directly print like in C language

awk 'BEGIN{for(i=32;i<=127;i++)printf "%c",i}';echo