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How to read string as hex number in bash?



2019 Community Moderator ElectionIn bash, how to convert 8 bytes to an unsigned int (64bit LE)?Bash: integer expression expected, using read/testHow to change bash prompt string in current bash session?How to convert an special hex character from an html page in bash?Loop over a string in zsh and BashModifying empty string variable inside switch statement in function bashExtract Complex String Inside Parentheses in Linux BashConstruct bash array with only string formatprintf escape %q string vs variableHow are ext4fs checksums being calculated?










6















I have the bash line:



expr substr $SUPERBLOCK 64 8


Which is return to me string line:



00080000


I know that this is, actually, a 0x00080000 in little-endian. Is there a way to create integer-variable from it in bash in big-endian like 0x80000?










share|improve this question




























    6















    I have the bash line:



    expr substr $SUPERBLOCK 64 8


    Which is return to me string line:



    00080000


    I know that this is, actually, a 0x00080000 in little-endian. Is there a way to create integer-variable from it in bash in big-endian like 0x80000?










    share|improve this question


























      6












      6








      6








      I have the bash line:



      expr substr $SUPERBLOCK 64 8


      Which is return to me string line:



      00080000


      I know that this is, actually, a 0x00080000 in little-endian. Is there a way to create integer-variable from it in bash in big-endian like 0x80000?










      share|improve this question
















      I have the bash line:



      expr substr $SUPERBLOCK 64 8


      Which is return to me string line:



      00080000


      I know that this is, actually, a 0x00080000 in little-endian. Is there a way to create integer-variable from it in bash in big-endian like 0x80000?







      bash numeric-data hex expr






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 18 at 18:52









      Jesse_b

      13.8k23471




      13.8k23471










      asked Mar 18 at 18:12









      DenisNovacDenisNovac

      438




      438




















          2 Answers
          2






          active

          oldest

          votes


















          8














          Probably a better way to do this but I've come up with this solution which converts the number to decimal and then back to hex (and manually adds the 0x):



          printf '0x%xn' "$((16#00080000))"


          Which you could write as:



          printf '0x%xn' "$((16#$(expr substr "$SUPERBLOCK" 64 8)))"





          share|improve this answer























          • Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

            – DenisNovac
            Mar 18 at 18:53







          • 5





            @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

            – Ped7g
            Mar 18 at 18:55












          • Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

            – DenisNovac
            Mar 18 at 19:01











          • @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

            – Jesse_b
            Mar 18 at 19:07












          • Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

            – DenisNovac
            Mar 18 at 19:12


















          5














          There are two more or less standard (and ancient) command-line unix tools that offer very easy ways to convert numbers between different bases:



          $ echo '16'; echo i; echo 00080000; echo p; | dc
          524288
          $ echo 'ibase=16'; echo 00080000; | bc
          524288


          For normal human use I very much prefer bc, but when writing a program that generates code, especially from a parser of some sort, a stack-based tool like dc may be easier to deal with (and indeed the original version of bc was a front-end parser for dc).






          share|improve this answer






















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8














            Probably a better way to do this but I've come up with this solution which converts the number to decimal and then back to hex (and manually adds the 0x):



            printf '0x%xn' "$((16#00080000))"


            Which you could write as:



            printf '0x%xn' "$((16#$(expr substr "$SUPERBLOCK" 64 8)))"





            share|improve this answer























            • Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

              – DenisNovac
              Mar 18 at 18:53







            • 5





              @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

              – Ped7g
              Mar 18 at 18:55












            • Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

              – DenisNovac
              Mar 18 at 19:01











            • @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

              – Jesse_b
              Mar 18 at 19:07












            • Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

              – DenisNovac
              Mar 18 at 19:12















            8














            Probably a better way to do this but I've come up with this solution which converts the number to decimal and then back to hex (and manually adds the 0x):



            printf '0x%xn' "$((16#00080000))"


            Which you could write as:



            printf '0x%xn' "$((16#$(expr substr "$SUPERBLOCK" 64 8)))"





            share|improve this answer























            • Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

              – DenisNovac
              Mar 18 at 18:53







            • 5





              @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

              – Ped7g
              Mar 18 at 18:55












            • Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

              – DenisNovac
              Mar 18 at 19:01











            • @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

              – Jesse_b
              Mar 18 at 19:07












            • Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

              – DenisNovac
              Mar 18 at 19:12













            8












            8








            8







            Probably a better way to do this but I've come up with this solution which converts the number to decimal and then back to hex (and manually adds the 0x):



            printf '0x%xn' "$((16#00080000))"


            Which you could write as:



            printf '0x%xn' "$((16#$(expr substr "$SUPERBLOCK" 64 8)))"





            share|improve this answer













            Probably a better way to do this but I've come up with this solution which converts the number to decimal and then back to hex (and manually adds the 0x):



            printf '0x%xn' "$((16#00080000))"


            Which you could write as:



            printf '0x%xn' "$((16#$(expr substr "$SUPERBLOCK" 64 8)))"






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 18 at 18:49









            Jesse_bJesse_b

            13.8k23471




            13.8k23471












            • Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

              – DenisNovac
              Mar 18 at 18:53







            • 5





              @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

              – Ped7g
              Mar 18 at 18:55












            • Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

              – DenisNovac
              Mar 18 at 19:01











            • @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

              – Jesse_b
              Mar 18 at 19:07












            • Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

              – DenisNovac
              Mar 18 at 19:12

















            • Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

              – DenisNovac
              Mar 18 at 18:53







            • 5





              @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

              – Ped7g
              Mar 18 at 18:55












            • Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

              – DenisNovac
              Mar 18 at 19:01











            • @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

              – Jesse_b
              Mar 18 at 19:07












            • Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

              – DenisNovac
              Mar 18 at 19:12
















            Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

            – DenisNovac
            Mar 18 at 18:53






            Thank you! I actually added |rev inside to convert to big-endian: printf "$((16#$(expr substr $SUPERBLOCK 64 8|rev)))"

            – DenisNovac
            Mar 18 at 18:53





            5




            5





            @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

            – Ped7g
            Mar 18 at 18:55






            @DenisNovac I'm not sure if you use big/little endian correctly (maybe you have something else on mind, but I'm doing some assembly programming for fun, so for me endianness is per bytes), but 0x12345678 is in other endianness 0x78563412, not 0x87654321. (and the value in your question 00080000 is after byte swap 00000800, i.e. 2048 decimal)

            – Ped7g
            Mar 18 at 18:55














            Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

            – DenisNovac
            Mar 18 at 19:01





            Oh, you are right. I just got the right answer by wrong way. I am rewriting some code from python to bash, so i know all answers before i got them.

            – DenisNovac
            Mar 18 at 19:01













            @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

            – Jesse_b
            Mar 18 at 19:07






            @DenisNovac: You didn't have the right answer FYI. 0x8000 was originally in your question which is not the same as 0x80000 or 0x00080000

            – Jesse_b
            Mar 18 at 19:07














            Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

            – DenisNovac
            Mar 18 at 19:12





            Yes, but i needed to get exactly 0x8000, so i've made mistake somewhere before. This is offset or something.

            – DenisNovac
            Mar 18 at 19:12













            5














            There are two more or less standard (and ancient) command-line unix tools that offer very easy ways to convert numbers between different bases:



            $ echo '16'; echo i; echo 00080000; echo p; | dc
            524288
            $ echo 'ibase=16'; echo 00080000; | bc
            524288


            For normal human use I very much prefer bc, but when writing a program that generates code, especially from a parser of some sort, a stack-based tool like dc may be easier to deal with (and indeed the original version of bc was a front-end parser for dc).






            share|improve this answer



























              5














              There are two more or less standard (and ancient) command-line unix tools that offer very easy ways to convert numbers between different bases:



              $ echo '16'; echo i; echo 00080000; echo p; | dc
              524288
              $ echo 'ibase=16'; echo 00080000; | bc
              524288


              For normal human use I very much prefer bc, but when writing a program that generates code, especially from a parser of some sort, a stack-based tool like dc may be easier to deal with (and indeed the original version of bc was a front-end parser for dc).






              share|improve this answer

























                5












                5








                5







                There are two more or less standard (and ancient) command-line unix tools that offer very easy ways to convert numbers between different bases:



                $ echo '16'; echo i; echo 00080000; echo p; | dc
                524288
                $ echo 'ibase=16'; echo 00080000; | bc
                524288


                For normal human use I very much prefer bc, but when writing a program that generates code, especially from a parser of some sort, a stack-based tool like dc may be easier to deal with (and indeed the original version of bc was a front-end parser for dc).






                share|improve this answer













                There are two more or less standard (and ancient) command-line unix tools that offer very easy ways to convert numbers between different bases:



                $ echo '16'; echo i; echo 00080000; echo p; | dc
                524288
                $ echo 'ibase=16'; echo 00080000; | bc
                524288


                For normal human use I very much prefer bc, but when writing a program that generates code, especially from a parser of some sort, a stack-based tool like dc may be easier to deal with (and indeed the original version of bc was a front-end parser for dc).







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 18 at 23:19









                Greg A. WoodsGreg A. Woods

                55248




                55248



























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