converting hex to decimal with bash





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0















I've seen some strange things.
I tried to convert hex to dec with bash Shell



I used very very simple command.



$ g_receiverBeforeToken=1158e460913d00000
$ echo $((16#$g_receiverBeforeToken))
1553255926290448384


As you guys know, this result should be '20000000000000000000'.



When I put in any other hex number, it was correct. But only 1553255926290448384 was weird.










share|improve this question

























  • So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

    – lurker
    Nov 23 '18 at 4:34




















0















I've seen some strange things.
I tried to convert hex to dec with bash Shell



I used very very simple command.



$ g_receiverBeforeToken=1158e460913d00000
$ echo $((16#$g_receiverBeforeToken))
1553255926290448384


As you guys know, this result should be '20000000000000000000'.



When I put in any other hex number, it was correct. But only 1553255926290448384 was weird.










share|improve this question

























  • So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

    – lurker
    Nov 23 '18 at 4:34
















0












0








0


0






I've seen some strange things.
I tried to convert hex to dec with bash Shell



I used very very simple command.



$ g_receiverBeforeToken=1158e460913d00000
$ echo $((16#$g_receiverBeforeToken))
1553255926290448384


As you guys know, this result should be '20000000000000000000'.



When I put in any other hex number, it was correct. But only 1553255926290448384 was weird.










share|improve this question
















I've seen some strange things.
I tried to convert hex to dec with bash Shell



I used very very simple command.



$ g_receiverBeforeToken=1158e460913d00000
$ echo $((16#$g_receiverBeforeToken))
1553255926290448384


As you guys know, this result should be '20000000000000000000'.



When I put in any other hex number, it was correct. But only 1553255926290448384 was weird.







bash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 4:43









jkdev

5,28553565




5,28553565










asked Nov 23 '18 at 4:22









JuneJune

216




216













  • So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

    – lurker
    Nov 23 '18 at 4:34





















  • So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

    – lurker
    Nov 23 '18 at 4:34



















So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

– lurker
Nov 23 '18 at 4:34







So you're saying, for example, 1553255926290448383 works fine? By the way, are you aware that 20000000000000000000 is greater than 2**64? The word size on your machine is probably 64 bits. 64 bits is 16 hexadecimal digits. 1158e460913d00000 is 17 hexadecimal digits.

– lurker
Nov 23 '18 at 4:34














2 Answers
2






active

oldest

votes


















1














It's not just that number, it's any number over 7fffffffffffffff, because it's using 64-bit integers and that's the largest one. 16-digit numbers over that wrap around and become negative (because of two's complement representation of signed integers):



$ echo $((16#7fffffffffffffff))
9223372036854775807
$ echo $((16#7fffffffffffffff + 1))
-9223372036854775808
$ echo $((16#8000000000000000))
-9223372036854775808


Past ffffffffffffffff (aka -1), it wraps back to zero:



$ echo $((16#ffffffffffffffff))
-1
$ echo $((16#ffffffffffffffff + 1))
0
$ echo $((16#10000000000000000))
0


Net result: only the last 16 hex digits actually matter; anything past that gets dropped off the high end of the 64-bit integer representation:



$ echo $((16#0000000000000010))
16
$ echo $((16#10000000000000010))
16
$ echo $((16#ffffffff0000000000000010))
16


Since 1553255926290448384 is 17 digits long, the first digit is being dropped off in this way:



$ echo $((16#1158e460913d00000))
1553255926290448384
$ echo $((16#158e460913d00000))
1553255926290448384
$ echo $((16#121345158e460913d00000))
1553255926290448384





share|improve this answer































    1














    You can do what you are looking to do using bc as a base conversion with a short script:



    #!/bin/bash

    ## validate sufficient input
    test -n "$1" || {
    printf "n error: insufficient input. usage: %s num [obase (2)] [ibase (10)]nn" "${0//*//}"
    exit 1
    }

    ## test for help
    test "$1" = "-h" || test "$1" = "--help" && {
    printf "n usage: %s num [obase (2)] [ibase (10)] -- to convert numbernn" "${0//*//}"
    exit 0
    }

    ## validate numeric value given for conversion (bash only test)
    ival="${1^^}"
    [[ $ival =~ [^0-9A-F] ]] && {
    printf "n error: invalid input. Input must be within upper/lower case hex character set [0-9A-Fa-f]nn"
    exit 1
    }

    ob=${2:-2}
    ib=${3:-10}

    # set obase first before ibase -- or weird things happen.
    printf "obase=%d; ibase=%d; %sn" $ob $ib $ival | bc


    Example Use/Output



    $ bash hex2dec.sh -h

    usage: hex2dec.sh num [obase (2)] [ibase (10)] -- to convert number


    Using your example:



    $ bash hex2dec.sh 1158e460913d00000 10 16
    20000000000000000000


    Also handy if you want it in binary as well:



    $ bash hex2dec.sh 1158e460913d00000 2 16
    10001010110001110010001100000100100010011110100000000000000000000





    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









      1














      It's not just that number, it's any number over 7fffffffffffffff, because it's using 64-bit integers and that's the largest one. 16-digit numbers over that wrap around and become negative (because of two's complement representation of signed integers):



      $ echo $((16#7fffffffffffffff))
      9223372036854775807
      $ echo $((16#7fffffffffffffff + 1))
      -9223372036854775808
      $ echo $((16#8000000000000000))
      -9223372036854775808


      Past ffffffffffffffff (aka -1), it wraps back to zero:



      $ echo $((16#ffffffffffffffff))
      -1
      $ echo $((16#ffffffffffffffff + 1))
      0
      $ echo $((16#10000000000000000))
      0


      Net result: only the last 16 hex digits actually matter; anything past that gets dropped off the high end of the 64-bit integer representation:



      $ echo $((16#0000000000000010))
      16
      $ echo $((16#10000000000000010))
      16
      $ echo $((16#ffffffff0000000000000010))
      16


      Since 1553255926290448384 is 17 digits long, the first digit is being dropped off in this way:



      $ echo $((16#1158e460913d00000))
      1553255926290448384
      $ echo $((16#158e460913d00000))
      1553255926290448384
      $ echo $((16#121345158e460913d00000))
      1553255926290448384





      share|improve this answer




























        1














        It's not just that number, it's any number over 7fffffffffffffff, because it's using 64-bit integers and that's the largest one. 16-digit numbers over that wrap around and become negative (because of two's complement representation of signed integers):



        $ echo $((16#7fffffffffffffff))
        9223372036854775807
        $ echo $((16#7fffffffffffffff + 1))
        -9223372036854775808
        $ echo $((16#8000000000000000))
        -9223372036854775808


        Past ffffffffffffffff (aka -1), it wraps back to zero:



        $ echo $((16#ffffffffffffffff))
        -1
        $ echo $((16#ffffffffffffffff + 1))
        0
        $ echo $((16#10000000000000000))
        0


        Net result: only the last 16 hex digits actually matter; anything past that gets dropped off the high end of the 64-bit integer representation:



        $ echo $((16#0000000000000010))
        16
        $ echo $((16#10000000000000010))
        16
        $ echo $((16#ffffffff0000000000000010))
        16


        Since 1553255926290448384 is 17 digits long, the first digit is being dropped off in this way:



        $ echo $((16#1158e460913d00000))
        1553255926290448384
        $ echo $((16#158e460913d00000))
        1553255926290448384
        $ echo $((16#121345158e460913d00000))
        1553255926290448384





        share|improve this answer


























          1












          1








          1







          It's not just that number, it's any number over 7fffffffffffffff, because it's using 64-bit integers and that's the largest one. 16-digit numbers over that wrap around and become negative (because of two's complement representation of signed integers):



          $ echo $((16#7fffffffffffffff))
          9223372036854775807
          $ echo $((16#7fffffffffffffff + 1))
          -9223372036854775808
          $ echo $((16#8000000000000000))
          -9223372036854775808


          Past ffffffffffffffff (aka -1), it wraps back to zero:



          $ echo $((16#ffffffffffffffff))
          -1
          $ echo $((16#ffffffffffffffff + 1))
          0
          $ echo $((16#10000000000000000))
          0


          Net result: only the last 16 hex digits actually matter; anything past that gets dropped off the high end of the 64-bit integer representation:



          $ echo $((16#0000000000000010))
          16
          $ echo $((16#10000000000000010))
          16
          $ echo $((16#ffffffff0000000000000010))
          16


          Since 1553255926290448384 is 17 digits long, the first digit is being dropped off in this way:



          $ echo $((16#1158e460913d00000))
          1553255926290448384
          $ echo $((16#158e460913d00000))
          1553255926290448384
          $ echo $((16#121345158e460913d00000))
          1553255926290448384





          share|improve this answer













          It's not just that number, it's any number over 7fffffffffffffff, because it's using 64-bit integers and that's the largest one. 16-digit numbers over that wrap around and become negative (because of two's complement representation of signed integers):



          $ echo $((16#7fffffffffffffff))
          9223372036854775807
          $ echo $((16#7fffffffffffffff + 1))
          -9223372036854775808
          $ echo $((16#8000000000000000))
          -9223372036854775808


          Past ffffffffffffffff (aka -1), it wraps back to zero:



          $ echo $((16#ffffffffffffffff))
          -1
          $ echo $((16#ffffffffffffffff + 1))
          0
          $ echo $((16#10000000000000000))
          0


          Net result: only the last 16 hex digits actually matter; anything past that gets dropped off the high end of the 64-bit integer representation:



          $ echo $((16#0000000000000010))
          16
          $ echo $((16#10000000000000010))
          16
          $ echo $((16#ffffffff0000000000000010))
          16


          Since 1553255926290448384 is 17 digits long, the first digit is being dropped off in this way:



          $ echo $((16#1158e460913d00000))
          1553255926290448384
          $ echo $((16#158e460913d00000))
          1553255926290448384
          $ echo $((16#121345158e460913d00000))
          1553255926290448384






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 23 '18 at 4:44









          Gordon DavissonGordon Davisson

          71.6k97994




          71.6k97994

























              1














              You can do what you are looking to do using bc as a base conversion with a short script:



              #!/bin/bash

              ## validate sufficient input
              test -n "$1" || {
              printf "n error: insufficient input. usage: %s num [obase (2)] [ibase (10)]nn" "${0//*//}"
              exit 1
              }

              ## test for help
              test "$1" = "-h" || test "$1" = "--help" && {
              printf "n usage: %s num [obase (2)] [ibase (10)] -- to convert numbernn" "${0//*//}"
              exit 0
              }

              ## validate numeric value given for conversion (bash only test)
              ival="${1^^}"
              [[ $ival =~ [^0-9A-F] ]] && {
              printf "n error: invalid input. Input must be within upper/lower case hex character set [0-9A-Fa-f]nn"
              exit 1
              }

              ob=${2:-2}
              ib=${3:-10}

              # set obase first before ibase -- or weird things happen.
              printf "obase=%d; ibase=%d; %sn" $ob $ib $ival | bc


              Example Use/Output



              $ bash hex2dec.sh -h

              usage: hex2dec.sh num [obase (2)] [ibase (10)] -- to convert number


              Using your example:



              $ bash hex2dec.sh 1158e460913d00000 10 16
              20000000000000000000


              Also handy if you want it in binary as well:



              $ bash hex2dec.sh 1158e460913d00000 2 16
              10001010110001110010001100000100100010011110100000000000000000000





              share|improve this answer






























                1














                You can do what you are looking to do using bc as a base conversion with a short script:



                #!/bin/bash

                ## validate sufficient input
                test -n "$1" || {
                printf "n error: insufficient input. usage: %s num [obase (2)] [ibase (10)]nn" "${0//*//}"
                exit 1
                }

                ## test for help
                test "$1" = "-h" || test "$1" = "--help" && {
                printf "n usage: %s num [obase (2)] [ibase (10)] -- to convert numbernn" "${0//*//}"
                exit 0
                }

                ## validate numeric value given for conversion (bash only test)
                ival="${1^^}"
                [[ $ival =~ [^0-9A-F] ]] && {
                printf "n error: invalid input. Input must be within upper/lower case hex character set [0-9A-Fa-f]nn"
                exit 1
                }

                ob=${2:-2}
                ib=${3:-10}

                # set obase first before ibase -- or weird things happen.
                printf "obase=%d; ibase=%d; %sn" $ob $ib $ival | bc


                Example Use/Output



                $ bash hex2dec.sh -h

                usage: hex2dec.sh num [obase (2)] [ibase (10)] -- to convert number


                Using your example:



                $ bash hex2dec.sh 1158e460913d00000 10 16
                20000000000000000000


                Also handy if you want it in binary as well:



                $ bash hex2dec.sh 1158e460913d00000 2 16
                10001010110001110010001100000100100010011110100000000000000000000





                share|improve this answer




























                  1












                  1








                  1







                  You can do what you are looking to do using bc as a base conversion with a short script:



                  #!/bin/bash

                  ## validate sufficient input
                  test -n "$1" || {
                  printf "n error: insufficient input. usage: %s num [obase (2)] [ibase (10)]nn" "${0//*//}"
                  exit 1
                  }

                  ## test for help
                  test "$1" = "-h" || test "$1" = "--help" && {
                  printf "n usage: %s num [obase (2)] [ibase (10)] -- to convert numbernn" "${0//*//}"
                  exit 0
                  }

                  ## validate numeric value given for conversion (bash only test)
                  ival="${1^^}"
                  [[ $ival =~ [^0-9A-F] ]] && {
                  printf "n error: invalid input. Input must be within upper/lower case hex character set [0-9A-Fa-f]nn"
                  exit 1
                  }

                  ob=${2:-2}
                  ib=${3:-10}

                  # set obase first before ibase -- or weird things happen.
                  printf "obase=%d; ibase=%d; %sn" $ob $ib $ival | bc


                  Example Use/Output



                  $ bash hex2dec.sh -h

                  usage: hex2dec.sh num [obase (2)] [ibase (10)] -- to convert number


                  Using your example:



                  $ bash hex2dec.sh 1158e460913d00000 10 16
                  20000000000000000000


                  Also handy if you want it in binary as well:



                  $ bash hex2dec.sh 1158e460913d00000 2 16
                  10001010110001110010001100000100100010011110100000000000000000000





                  share|improve this answer















                  You can do what you are looking to do using bc as a base conversion with a short script:



                  #!/bin/bash

                  ## validate sufficient input
                  test -n "$1" || {
                  printf "n error: insufficient input. usage: %s num [obase (2)] [ibase (10)]nn" "${0//*//}"
                  exit 1
                  }

                  ## test for help
                  test "$1" = "-h" || test "$1" = "--help" && {
                  printf "n usage: %s num [obase (2)] [ibase (10)] -- to convert numbernn" "${0//*//}"
                  exit 0
                  }

                  ## validate numeric value given for conversion (bash only test)
                  ival="${1^^}"
                  [[ $ival =~ [^0-9A-F] ]] && {
                  printf "n error: invalid input. Input must be within upper/lower case hex character set [0-9A-Fa-f]nn"
                  exit 1
                  }

                  ob=${2:-2}
                  ib=${3:-10}

                  # set obase first before ibase -- or weird things happen.
                  printf "obase=%d; ibase=%d; %sn" $ob $ib $ival | bc


                  Example Use/Output



                  $ bash hex2dec.sh -h

                  usage: hex2dec.sh num [obase (2)] [ibase (10)] -- to convert number


                  Using your example:



                  $ bash hex2dec.sh 1158e460913d00000 10 16
                  20000000000000000000


                  Also handy if you want it in binary as well:



                  $ bash hex2dec.sh 1158e460913d00000 2 16
                  10001010110001110010001100000100100010011110100000000000000000000






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 23 '18 at 4:56

























                  answered Nov 23 '18 at 4:50









                  David C. RankinDavid C. Rankin

                  43.9k33252




                  43.9k33252






























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