I am looking for hexadecimal increment by reading user input using shell script. For example the user input is 01A. I want to increase the hex count by 4 times
Looking for output:
01A
O1B
01C
01D
01E
Can someone suggest?
to start with:
echo 'obase=ibase=16;1A+1' |bc
Hi sai_1712,
One way using perl:
$ perl -e '$var = 0x01A; for ( 1 .. 5 ) { printf qq[%03X\n], $var++ }'
01A
01B
01C
01D
01E
Regards,
Birei
Thank you vgersh99 & birei for suggestion...
I am new to shell scripting. This hexadecimal increment count is part of my customized script. I get the output like below when i run my script
Cat lun.txt
01A
01A
The line1 & line2 values in lun.txt are same.The lun.txt output varies when ever I run my script and the increment count also varies it may be 2 or 3 or 4 depend upon requirement
I want to call the line1 variable(hexadecimal) from lun.txt file and increment by 2 or 3 or 4......... times which will define by user.
Your suggestions will be more helpful to complete my customized script.
Please note that I've written my script in shell
#!/bin/ksh
incr=3
while read val
do
echo "obase=ibase=16;${val}+${incr}" |bc
done < lun.txt
Using bash:
#!/bin/bash
incr=4
while read val
do
printf "%X\n" $((0x${val}+incr))
done < lun.txt
or perhaps your after:
read -p "Enter increment: " inc
read -p "Enter HEX value: " hex
printf "HEX + increment is: %X\n" $((0x$hex+inc))
Hello vgersh99,
When I use the below code I am getting the below output.
#!/bin/bash
incr=3
while read val
do
echo "obase=ibase=16;${val}+${incr}" |bc
done < lun.txt
Output:
1D
***********************
lun.txt file contains
01A
***********************
I am looking for the below output. The increment is variable and to be defined by the user every time depend on requirement
01A
01B
01C
01D
---------- Post updated at 04:06 PM ---------- Previous update was at 04:00 PM ----------
Hello Chubler,
The below code is not working. Please let me know where I've to modify.The increment is a variable and to be defined by the user every time depend on requirement
#!/bin/bash
read -p "Enter increment: " inc
read -p "Enter HEX value: " hex
printf "HEX + increment is: %X\n" $((0x$hex+inc))
#!/bin/bash
incr=4
while read val
do
printf "%X\n" $((0x${val}+incr))
done < lun.txt
Output:
syntax error at line 5: `(' unexpected
#!/bin/bash
read -p "Enter increment: " inc
read -p "Enter HEX value: " hex
printf "HEX + increment is: %X\n" $((0x$hex+inc))
Output:
-p: is not an identifier
Building on the above code it became necessary to test the script for a larger number of numbers because the arithmetic in the "bc" is all in hexadecimal. The "bc" command expects upper-case hexadecimal. Some adjustment to output to put the leading zero on your hex output (which is unconventional).
max=20 # Different numbers to generate
while read val
do
incr=0
while [ ${incr} -le ${max} ]
do
# Convert $incr to upper case hexadecimal for "bc"
incrx=$(printf "%#x\n" ${incr}|sed -e "s/x//g"|tr '[a-z]' '[A-Z]')
hex=$(echo "obase=ibase=16;${val}+${incrx}" |bc)
printf "0${hex}\n"
incr=$((${incr} + 1))
done
done < lun.txt
01A
01B
01C
01D
01E
01F
020
021
022
023
024
025
026
027
028
029
02A
02B
02C
02D
02E
This code didn't execute. Got the below error..
#! /bin/bash
max=20 # Different numbers to generate
while read val
do
incr=0
while [ ${incr} -le ${max} ]
do
# Convert $incr to upper case hexadecimal for "bc"
incrx=$(printf "%#x\n" ${incr}|sed -e "s/x//g"|tr '[a-z]' '[A-Z]')
hex=$(echo "obase=ibase=16;${val}+${incrx}" |bc)
printf "0${hex}\n"
incr=$((${incr} + 1))
done
done < lun.txt
syntax error at line 9: `incrx=$' unexpected
Since read -p "prompt" var and $() dont work with your version of bash it must be quite old.
What OS are you on? Can you show us the output of:
bash -c 'echo $BASH_VERSION'
on your system?
2.03.0(1)-release
Unmodernised version for Bourne Shell of the day:
max=20 # Different numbers to generate
cat lun.txt | while read val
do
incr=0
while [ ${incr} -le ${max} ]
do
# Convert $incr to upper case hexadecimal for "bc"
incrx=`printf "%#x\n" ${incr}|sed -e "s/x//g"|tr '[a-z]' '[A-Z]'`
hex=`echo "obase=ibase=16;${val}+${incrx}" |bc`
printf "0${hex}\n"
incr=`expr ${incr} + 1`
done
done
Yes I have noticed the "cat".
The above code is working... Thank you all for your help!!!!!!!
Wow! 2.03 is over 12 years out of date, it might even have Y2K issues 
At last I cannot be criticised for posting perfectly normal Shell script, complete with backticks, "expr" and "cat". This used to be quite normal in system-supplied startup scripts.