BASH base conversion from decimal to hex












21














In Bash, how does one do base conversion from decimal to another base, especially hex. It seems easy to go the other way:



$ echo $((16#55))
85


With a web-search, I found a script that does the maths and character manipulation to do the conversion, and I could use that as a function, but I'd have thought that bash would already have a built-in base conversion -- does it?










share|improve this question






















  • stackoverflow.com/questions/378829/…
    – Ciro Santilli 新疆改造中心 六四事件 法轮功
    Apr 20 '17 at 12:48
















21














In Bash, how does one do base conversion from decimal to another base, especially hex. It seems easy to go the other way:



$ echo $((16#55))
85


With a web-search, I found a script that does the maths and character manipulation to do the conversion, and I could use that as a function, but I'd have thought that bash would already have a built-in base conversion -- does it?










share|improve this question






















  • stackoverflow.com/questions/378829/…
    – Ciro Santilli 新疆改造中心 六四事件 法轮功
    Apr 20 '17 at 12:48














21












21








21


6





In Bash, how does one do base conversion from decimal to another base, especially hex. It seems easy to go the other way:



$ echo $((16#55))
85


With a web-search, I found a script that does the maths and character manipulation to do the conversion, and I could use that as a function, but I'd have thought that bash would already have a built-in base conversion -- does it?










share|improve this question













In Bash, how does one do base conversion from decimal to another base, especially hex. It seems easy to go the other way:



$ echo $((16#55))
85


With a web-search, I found a script that does the maths and character manipulation to do the conversion, and I could use that as a function, but I'd have thought that bash would already have a built-in base conversion -- does it?







bash shell text-processing hex






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 19 '15 at 13:19









Dave Rove

140116




140116












  • stackoverflow.com/questions/378829/…
    – Ciro Santilli 新疆改造中心 六四事件 法轮功
    Apr 20 '17 at 12:48


















  • stackoverflow.com/questions/378829/…
    – Ciro Santilli 新疆改造中心 六四事件 法轮功
    Apr 20 '17 at 12:48
















stackoverflow.com/questions/378829/…
– Ciro Santilli 新疆改造中心 六四事件 法轮功
Apr 20 '17 at 12:48




stackoverflow.com/questions/378829/…
– Ciro Santilli 新疆改造中心 六四事件 法轮功
Apr 20 '17 at 12:48










4 Answers
4






active

oldest

votes


















30














With bash (or any shell, provided the printf command is available (a standard POSIX command often built in the shells)):



printf '%xn' 85


​​​​​​​​​​​​​​​​​
With zsh, you can also do:



dec=85
hex=$(([##16]dec))


That works for bases from 2 to 36 (with 0-9a-z case insensitive as the digits).



With ksh93, you can use:



dec=85
base54=$(printf %..54 "$dec")


Which works for bases from 2 to 64 (with 0-9a-zA-Z@_ as the digits).



With ksh and zsh, there's also:



$ typeset -i34 x=123; echo "$x"
34#3l


Though that's limited to bases up to 36 in ksh88, zsh and pdksh and 64 in ksh93.



Note that all those are limited to the size of the long integers on your system (int's with some shells). For anything bigger, you can use bc or dc.



$ echo 'obase=16; 9999999999999999999999' | bc
21E19E0C9BAB23FFFFF
$ echo '16o 9999999999999999999999 p' | dc
21E19E0C9BAB23FFFFF


With supported bases ranging from 2 to some number required by POSIX to be at least as high as 99. For bases greater than 16, digits greater than 9 are represented as space-separated 0-padded decimal numbers.



$ echo 'obase=30; 123456' | bc
04 17 05 06


Or same with dc (bc used to be (and still is on some systems) a wrapper around dc):



$ echo 30o123456p | dc
04 17 05 06





share|improve this answer























  • Thanks. Exactly what I was looking for. (And embarrassingly simple.)
    – Dave Rove
    Mar 19 '15 at 13:32










  • What if you want to do an arbitrary base like you can specify with #?
    – flarn2006
    Sep 27 '16 at 18:29










  • @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
    – Stéphane Chazelas
    Sep 27 '16 at 20:27






  • 1




    @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
    – flarn2006
    Sep 28 '16 at 1:11



















5














Use printf:



$ printf "%d %xn" $((16#55)) $((10#85))
85 55


To assign the value to a variable use command substitution:



$ x=$( printf "%x" 85 ) ; echo $x
55





share|improve this answer

















  • 4




    Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
    – glenn jackman
    Mar 19 '15 at 13:37












  • @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
    – Janis
    Mar 19 '15 at 13:40






  • 3




    -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
    – Stéphane Chazelas
    Mar 19 '15 at 13:42












  • @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
    – glenn jackman
    Mar 19 '15 at 13:44








  • 1




    @glennjackman, yes, it only forks for executing external commands.
    – Stéphane Chazelas
    Mar 19 '15 at 13:45



















1














Use the built-in Arithmetic Expansion substitution present in all POSIX compliant shells - which is pretty much universal these days.



$ echo $((0xbc))
188


and



$ hex=dead
$ dec=$((0x$hex))
$ echo $dec
57005


CAUTION: Particularly in the last example the expansion could cause unexpected results - the hex digits in the variable 'hex' have to form a legal hex constant, otherwise potentially obscure error messages happen.
e.g. if 'hex' were '0xdead', the arithmetic expansion would become 0x0xdead, which is not interpretable as a constant. Of course, in that case the arithmetic expansion $(($hex)) would do the trick. It is left as an exercise for the reader to create the simple substring processing pattern matching that would remove an optional '0x' prefix.






share|improve this answer



















  • 2




    You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
    – Thomas Guyot-Sionnest
    Feb 9 '16 at 6:48



















0














You can use the awk Velour library:



$ velour -n 'print n_baseconv(15, 10, 16)'
F


Or:



$ velour -n 'print n_baseconv(ARGV[1], 10, 16)' 15
F





share|improve this answer























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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    30














    With bash (or any shell, provided the printf command is available (a standard POSIX command often built in the shells)):



    printf '%xn' 85


    ​​​​​​​​​​​​​​​​​
    With zsh, you can also do:



    dec=85
    hex=$(([##16]dec))


    That works for bases from 2 to 36 (with 0-9a-z case insensitive as the digits).



    With ksh93, you can use:



    dec=85
    base54=$(printf %..54 "$dec")


    Which works for bases from 2 to 64 (with 0-9a-zA-Z@_ as the digits).



    With ksh and zsh, there's also:



    $ typeset -i34 x=123; echo "$x"
    34#3l


    Though that's limited to bases up to 36 in ksh88, zsh and pdksh and 64 in ksh93.



    Note that all those are limited to the size of the long integers on your system (int's with some shells). For anything bigger, you can use bc or dc.



    $ echo 'obase=16; 9999999999999999999999' | bc
    21E19E0C9BAB23FFFFF
    $ echo '16o 9999999999999999999999 p' | dc
    21E19E0C9BAB23FFFFF


    With supported bases ranging from 2 to some number required by POSIX to be at least as high as 99. For bases greater than 16, digits greater than 9 are represented as space-separated 0-padded decimal numbers.



    $ echo 'obase=30; 123456' | bc
    04 17 05 06


    Or same with dc (bc used to be (and still is on some systems) a wrapper around dc):



    $ echo 30o123456p | dc
    04 17 05 06





    share|improve this answer























    • Thanks. Exactly what I was looking for. (And embarrassingly simple.)
      – Dave Rove
      Mar 19 '15 at 13:32










    • What if you want to do an arbitrary base like you can specify with #?
      – flarn2006
      Sep 27 '16 at 18:29










    • @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
      – Stéphane Chazelas
      Sep 27 '16 at 20:27






    • 1




      @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
      – flarn2006
      Sep 28 '16 at 1:11
















    30














    With bash (or any shell, provided the printf command is available (a standard POSIX command often built in the shells)):



    printf '%xn' 85


    ​​​​​​​​​​​​​​​​​
    With zsh, you can also do:



    dec=85
    hex=$(([##16]dec))


    That works for bases from 2 to 36 (with 0-9a-z case insensitive as the digits).



    With ksh93, you can use:



    dec=85
    base54=$(printf %..54 "$dec")


    Which works for bases from 2 to 64 (with 0-9a-zA-Z@_ as the digits).



    With ksh and zsh, there's also:



    $ typeset -i34 x=123; echo "$x"
    34#3l


    Though that's limited to bases up to 36 in ksh88, zsh and pdksh and 64 in ksh93.



    Note that all those are limited to the size of the long integers on your system (int's with some shells). For anything bigger, you can use bc or dc.



    $ echo 'obase=16; 9999999999999999999999' | bc
    21E19E0C9BAB23FFFFF
    $ echo '16o 9999999999999999999999 p' | dc
    21E19E0C9BAB23FFFFF


    With supported bases ranging from 2 to some number required by POSIX to be at least as high as 99. For bases greater than 16, digits greater than 9 are represented as space-separated 0-padded decimal numbers.



    $ echo 'obase=30; 123456' | bc
    04 17 05 06


    Or same with dc (bc used to be (and still is on some systems) a wrapper around dc):



    $ echo 30o123456p | dc
    04 17 05 06





    share|improve this answer























    • Thanks. Exactly what I was looking for. (And embarrassingly simple.)
      – Dave Rove
      Mar 19 '15 at 13:32










    • What if you want to do an arbitrary base like you can specify with #?
      – flarn2006
      Sep 27 '16 at 18:29










    • @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
      – Stéphane Chazelas
      Sep 27 '16 at 20:27






    • 1




      @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
      – flarn2006
      Sep 28 '16 at 1:11














    30












    30








    30






    With bash (or any shell, provided the printf command is available (a standard POSIX command often built in the shells)):



    printf '%xn' 85


    ​​​​​​​​​​​​​​​​​
    With zsh, you can also do:



    dec=85
    hex=$(([##16]dec))


    That works for bases from 2 to 36 (with 0-9a-z case insensitive as the digits).



    With ksh93, you can use:



    dec=85
    base54=$(printf %..54 "$dec")


    Which works for bases from 2 to 64 (with 0-9a-zA-Z@_ as the digits).



    With ksh and zsh, there's also:



    $ typeset -i34 x=123; echo "$x"
    34#3l


    Though that's limited to bases up to 36 in ksh88, zsh and pdksh and 64 in ksh93.



    Note that all those are limited to the size of the long integers on your system (int's with some shells). For anything bigger, you can use bc or dc.



    $ echo 'obase=16; 9999999999999999999999' | bc
    21E19E0C9BAB23FFFFF
    $ echo '16o 9999999999999999999999 p' | dc
    21E19E0C9BAB23FFFFF


    With supported bases ranging from 2 to some number required by POSIX to be at least as high as 99. For bases greater than 16, digits greater than 9 are represented as space-separated 0-padded decimal numbers.



    $ echo 'obase=30; 123456' | bc
    04 17 05 06


    Or same with dc (bc used to be (and still is on some systems) a wrapper around dc):



    $ echo 30o123456p | dc
    04 17 05 06





    share|improve this answer














    With bash (or any shell, provided the printf command is available (a standard POSIX command often built in the shells)):



    printf '%xn' 85


    ​​​​​​​​​​​​​​​​​
    With zsh, you can also do:



    dec=85
    hex=$(([##16]dec))


    That works for bases from 2 to 36 (with 0-9a-z case insensitive as the digits).



    With ksh93, you can use:



    dec=85
    base54=$(printf %..54 "$dec")


    Which works for bases from 2 to 64 (with 0-9a-zA-Z@_ as the digits).



    With ksh and zsh, there's also:



    $ typeset -i34 x=123; echo "$x"
    34#3l


    Though that's limited to bases up to 36 in ksh88, zsh and pdksh and 64 in ksh93.



    Note that all those are limited to the size of the long integers on your system (int's with some shells). For anything bigger, you can use bc or dc.



    $ echo 'obase=16; 9999999999999999999999' | bc
    21E19E0C9BAB23FFFFF
    $ echo '16o 9999999999999999999999 p' | dc
    21E19E0C9BAB23FFFFF


    With supported bases ranging from 2 to some number required by POSIX to be at least as high as 99. For bases greater than 16, digits greater than 9 are represented as space-separated 0-padded decimal numbers.



    $ echo 'obase=30; 123456' | bc
    04 17 05 06


    Or same with dc (bc used to be (and still is on some systems) a wrapper around dc):



    $ echo 30o123456p | dc
    04 17 05 06






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 11 hours ago

























    answered Mar 19 '15 at 13:30









    Stéphane Chazelas

    298k54563910




    298k54563910












    • Thanks. Exactly what I was looking for. (And embarrassingly simple.)
      – Dave Rove
      Mar 19 '15 at 13:32










    • What if you want to do an arbitrary base like you can specify with #?
      – flarn2006
      Sep 27 '16 at 18:29










    • @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
      – Stéphane Chazelas
      Sep 27 '16 at 20:27






    • 1




      @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
      – flarn2006
      Sep 28 '16 at 1:11


















    • Thanks. Exactly what I was looking for. (And embarrassingly simple.)
      – Dave Rove
      Mar 19 '15 at 13:32










    • What if you want to do an arbitrary base like you can specify with #?
      – flarn2006
      Sep 27 '16 at 18:29










    • @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
      – Stéphane Chazelas
      Sep 27 '16 at 20:27






    • 1




      @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
      – flarn2006
      Sep 28 '16 at 1:11
















    Thanks. Exactly what I was looking for. (And embarrassingly simple.)
    – Dave Rove
    Mar 19 '15 at 13:32




    Thanks. Exactly what I was looking for. (And embarrassingly simple.)
    – Dave Rove
    Mar 19 '15 at 13:32












    What if you want to do an arbitrary base like you can specify with #?
    – flarn2006
    Sep 27 '16 at 18:29




    What if you want to do an arbitrary base like you can specify with #?
    – flarn2006
    Sep 27 '16 at 18:29












    @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
    – Stéphane Chazelas
    Sep 27 '16 at 20:27




    @flarn2006. With bash builtins, you can input numbers in any base, but not output in any base other than 8, 10 and 16. For other bases, you'd need another shell like zsh or ksh or use bc/dc.
    – Stéphane Chazelas
    Sep 27 '16 at 20:27




    1




    1




    @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
    – flarn2006
    Sep 28 '16 at 1:11




    @StéphaneChazelas Really? That's kind of weird. It almost seems like they were too lazy to program in a syntax for it or something; there's no way the idea of outputting in any base didn't cross their minds if they implemented inputting.
    – flarn2006
    Sep 28 '16 at 1:11













    5














    Use printf:



    $ printf "%d %xn" $((16#55)) $((10#85))
    85 55


    To assign the value to a variable use command substitution:



    $ x=$( printf "%x" 85 ) ; echo $x
    55





    share|improve this answer

















    • 4




      Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
      – glenn jackman
      Mar 19 '15 at 13:37












    • @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
      – Janis
      Mar 19 '15 at 13:40






    • 3




      -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
      – Stéphane Chazelas
      Mar 19 '15 at 13:42












    • @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
      – glenn jackman
      Mar 19 '15 at 13:44








    • 1




      @glennjackman, yes, it only forks for executing external commands.
      – Stéphane Chazelas
      Mar 19 '15 at 13:45
















    5














    Use printf:



    $ printf "%d %xn" $((16#55)) $((10#85))
    85 55


    To assign the value to a variable use command substitution:



    $ x=$( printf "%x" 85 ) ; echo $x
    55





    share|improve this answer

















    • 4




      Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
      – glenn jackman
      Mar 19 '15 at 13:37












    • @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
      – Janis
      Mar 19 '15 at 13:40






    • 3




      -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
      – Stéphane Chazelas
      Mar 19 '15 at 13:42












    • @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
      – glenn jackman
      Mar 19 '15 at 13:44








    • 1




      @glennjackman, yes, it only forks for executing external commands.
      – Stéphane Chazelas
      Mar 19 '15 at 13:45














    5












    5








    5






    Use printf:



    $ printf "%d %xn" $((16#55)) $((10#85))
    85 55


    To assign the value to a variable use command substitution:



    $ x=$( printf "%x" 85 ) ; echo $x
    55





    share|improve this answer












    Use printf:



    $ printf "%d %xn" $((16#55)) $((10#85))
    85 55


    To assign the value to a variable use command substitution:



    $ x=$( printf "%x" 85 ) ; echo $x
    55






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Mar 19 '15 at 13:33









    Janis

    10.1k11437




    10.1k11437








    • 4




      Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
      – glenn jackman
      Mar 19 '15 at 13:37












    • @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
      – Janis
      Mar 19 '15 at 13:40






    • 3




      -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
      – Stéphane Chazelas
      Mar 19 '15 at 13:42












    • @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
      – glenn jackman
      Mar 19 '15 at 13:44








    • 1




      @glennjackman, yes, it only forks for executing external commands.
      – Stéphane Chazelas
      Mar 19 '15 at 13:45














    • 4




      Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
      – glenn jackman
      Mar 19 '15 at 13:37












    • @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
      – Janis
      Mar 19 '15 at 13:40






    • 3




      -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
      – Stéphane Chazelas
      Mar 19 '15 at 13:42












    • @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
      – glenn jackman
      Mar 19 '15 at 13:44








    • 1




      @glennjackman, yes, it only forks for executing external commands.
      – Stéphane Chazelas
      Mar 19 '15 at 13:45








    4




    4




    Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
    – glenn jackman
    Mar 19 '15 at 13:37






    Or, use the -v option for the builtin printf: printf -v foo "%d" $((16#BEEF)); echo $foo
    – glenn jackman
    Mar 19 '15 at 13:37














    @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
    – Janis
    Mar 19 '15 at 13:40




    @Glenn Using non-standard options (like -v) is unnecessary here; I'd avoid it.
    – Janis
    Mar 19 '15 at 13:40




    3




    3




    -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
    – Stéphane Chazelas
    Mar 19 '15 at 13:42






    -v would avoid the fork and pipe (in bash, not ksh93 which wouldn't fork here as printf is builtin). Note that $((16#55)) is not standard either.
    – Stéphane Chazelas
    Mar 19 '15 at 13:42














    @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
    – glenn jackman
    Mar 19 '15 at 13:44






    @StéphaneChazelas, wait a minute. Are you saying that ksh93 does not fork for x=$(some_builtin) ?
    – glenn jackman
    Mar 19 '15 at 13:44






    1




    1




    @glennjackman, yes, it only forks for executing external commands.
    – Stéphane Chazelas
    Mar 19 '15 at 13:45




    @glennjackman, yes, it only forks for executing external commands.
    – Stéphane Chazelas
    Mar 19 '15 at 13:45











    1














    Use the built-in Arithmetic Expansion substitution present in all POSIX compliant shells - which is pretty much universal these days.



    $ echo $((0xbc))
    188


    and



    $ hex=dead
    $ dec=$((0x$hex))
    $ echo $dec
    57005


    CAUTION: Particularly in the last example the expansion could cause unexpected results - the hex digits in the variable 'hex' have to form a legal hex constant, otherwise potentially obscure error messages happen.
    e.g. if 'hex' were '0xdead', the arithmetic expansion would become 0x0xdead, which is not interpretable as a constant. Of course, in that case the arithmetic expansion $(($hex)) would do the trick. It is left as an exercise for the reader to create the simple substring processing pattern matching that would remove an optional '0x' prefix.






    share|improve this answer



















    • 2




      You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
      – Thomas Guyot-Sionnest
      Feb 9 '16 at 6:48
















    1














    Use the built-in Arithmetic Expansion substitution present in all POSIX compliant shells - which is pretty much universal these days.



    $ echo $((0xbc))
    188


    and



    $ hex=dead
    $ dec=$((0x$hex))
    $ echo $dec
    57005


    CAUTION: Particularly in the last example the expansion could cause unexpected results - the hex digits in the variable 'hex' have to form a legal hex constant, otherwise potentially obscure error messages happen.
    e.g. if 'hex' were '0xdead', the arithmetic expansion would become 0x0xdead, which is not interpretable as a constant. Of course, in that case the arithmetic expansion $(($hex)) would do the trick. It is left as an exercise for the reader to create the simple substring processing pattern matching that would remove an optional '0x' prefix.






    share|improve this answer



















    • 2




      You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
      – Thomas Guyot-Sionnest
      Feb 9 '16 at 6:48














    1












    1








    1






    Use the built-in Arithmetic Expansion substitution present in all POSIX compliant shells - which is pretty much universal these days.



    $ echo $((0xbc))
    188


    and



    $ hex=dead
    $ dec=$((0x$hex))
    $ echo $dec
    57005


    CAUTION: Particularly in the last example the expansion could cause unexpected results - the hex digits in the variable 'hex' have to form a legal hex constant, otherwise potentially obscure error messages happen.
    e.g. if 'hex' were '0xdead', the arithmetic expansion would become 0x0xdead, which is not interpretable as a constant. Of course, in that case the arithmetic expansion $(($hex)) would do the trick. It is left as an exercise for the reader to create the simple substring processing pattern matching that would remove an optional '0x' prefix.






    share|improve this answer














    Use the built-in Arithmetic Expansion substitution present in all POSIX compliant shells - which is pretty much universal these days.



    $ echo $((0xbc))
    188


    and



    $ hex=dead
    $ dec=$((0x$hex))
    $ echo $dec
    57005


    CAUTION: Particularly in the last example the expansion could cause unexpected results - the hex digits in the variable 'hex' have to form a legal hex constant, otherwise potentially obscure error messages happen.
    e.g. if 'hex' were '0xdead', the arithmetic expansion would become 0x0xdead, which is not interpretable as a constant. Of course, in that case the arithmetic expansion $(($hex)) would do the trick. It is left as an exercise for the reader to create the simple substring processing pattern matching that would remove an optional '0x' prefix.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 29 '15 at 23:58

























    answered Dec 29 '15 at 23:42









    Mark van der Pol

    272




    272








    • 2




      You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
      – Thomas Guyot-Sionnest
      Feb 9 '16 at 6:48














    • 2




      You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
      – Thomas Guyot-Sionnest
      Feb 9 '16 at 6:48








    2




    2




    You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
    – Thomas Guyot-Sionnest
    Feb 9 '16 at 6:48




    You're doing the exact same thing the OP did in his example; he's looking for the reverse operation.
    – Thomas Guyot-Sionnest
    Feb 9 '16 at 6:48











    0














    You can use the awk Velour library:



    $ velour -n 'print n_baseconv(15, 10, 16)'
    F


    Or:



    $ velour -n 'print n_baseconv(ARGV[1], 10, 16)' 15
    F





    share|improve this answer




























      0














      You can use the awk Velour library:



      $ velour -n 'print n_baseconv(15, 10, 16)'
      F


      Or:



      $ velour -n 'print n_baseconv(ARGV[1], 10, 16)' 15
      F





      share|improve this answer


























        0












        0








        0






        You can use the awk Velour library:



        $ velour -n 'print n_baseconv(15, 10, 16)'
        F


        Or:



        $ velour -n 'print n_baseconv(ARGV[1], 10, 16)' 15
        F





        share|improve this answer














        You can use the awk Velour library:



        $ velour -n 'print n_baseconv(15, 10, 16)'
        F


        Or:



        $ velour -n 'print n_baseconv(ARGV[1], 10, 16)' 15
        F






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jul 1 at 14:10

























        answered Mar 26 '17 at 23:12









        Steven Penny

        1




        1






























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