bash convert hex string to integer code using printf failing

Question

I am iterating over hex digits in a string using a for loop. I can extract each hex digit ok, but my code to convert to a number is giving strange values. How can I fix this code?

Script:

#!/bin/bash

mystring="5e51584a4c"

for (( i = 0; i < ${#mystring}; i = i + 2)); do
    snumber="${mystring:i:2}"
    printf "number as string=%s\n" $snumber
    number=$(printf "%x"  "'${mystring:i:2}")
    printf "number=%d\n" $number
done

I am getting this output:

number as string=5e
number=35
number as string=51
number=35
number as string=58
number=35
number as string=4a
number=34
number as string=4c
number=34

oguz ismail · Accepted Answer · 2020-06-19 05:12:09Z

5

You don't need printf in this case, you can use bash's $((base#number)) construct.

#!/bin/bash -

mystring="5e51584a4c"

for ((i=0; i<${#mystring}; i+=2)); do
    snumber="${mystring:i:2}"
    echo "number as string=${snumber}"
    echo "number=$((16#${snumber}))"
done

edited Jun 19, 2020 at 5:12

answered Dec 7, 2018 at 11:31

oguz ismail

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Comments

Maxim Egorushkin · Accepted Answer · 2018-12-07 10:21:22Z

4

Hex numbers need to have 0x prefix:

printf "%d\n" 0x5e

Outputs:

mystring="5e51584a4c"

for (( i = 0; i < ${#mystring}; i = i + 2)); do
    snumber="${mystring:i:2}"
    printf "number=%d\n" 0x$snumber
done

answered Dec 7, 2018 at 10:21

Maxim Egorushkin

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Comments

Toby Speight · Accepted Answer · 2018-12-07 12:01:25Z

3

Your printf command is printing the character code of the first digit in each substring, because you have the ' prefix to the argument. You want a 0x prefix instead.

When printing as hexadecimal, you can use %#x conversion to make printf emit the leading 0x prefix needed by the next usage:

number=$(printf '%#x'  "0x${mystring:i:2}")
#                ^^^    ^^

answered Dec 7, 2018 at 12:01

Toby Speight

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