I just assigned a variable but echo variable shows something else

Question

Here are a series of cases where echo  var can show a different value than what was just assigned  This happens regardless of whether the assigned value was  double quoted    single quoted  or unquoted    How do I get the shell to set my variable correctly   Asterisks  The expected output is    Foobar is free software     but instead I get a list of filenames     var     Foobar is free software       echo  var   bin  boot  dev  etc  home  initrd img  lib  lib64  media  mnt  opt  proc       Square brackets  The expected value is  a-z   but sometimes I get a single letter instead     var  a-z    echo  var c   Line feeds  newlines   The expected value is a  a list of separate lines  but instead all the values are on one line     cat file foo bar baz    var   cat file    echo  var foo bar baz   Multiple spaces  I expected a carefully aligned table header  but instead multiple spaces either disappear or are collapsed into one     var         title          count    echo  var title   count   Tabs  I expected two tab separated values  but instead I get two space separated values     var   key tvalue    echo  var key value

User · Answer

user double quote to get the exact value  like this   echo    var     and it will read your value correctly

User · Answer

Additional to putting the variable in quotation  one could also translate the output of the variable using tr and converting spaces to newlines     echo  var   tr       n  foo bar baz   Although this is a little more convoluted  it does add more diversity with the output as you can substitute any character as the separator between array variables

User · Answer

In addition to other issues caused by failing to quote  -n and -e can be consumed by echo as arguments   Only the former is legal per the POSIX spec for echo  but several common implementations violate the spec and consume -e as well    To avoid this  use printf instead of echo when details matter   Thus     vars  -e -n -a    echo  vars        breaks because -e and -n can be treated as arguments to echo -a   echo   vars  -e -n -a   However  correct quoting won t always save you when using echo     vars  -n    echo  vars      not even an empty line was printed      whereas it will save you with printf     vars  -n    printf   s n    vars  -n

User · Answer

echo  var output highly depends on the value of IFS variable  By default it contains space  tab  and newline characters    ks localhost     echo -n   IFS    cat -vte   I    This means that when shell is doing field splitting  or word splitting  it uses all these characters as word separators  This is what happens when referencing a variable without double quotes to echo it   var  and thus expected output is altered   One way to prevent word splitting  besides using double quotes  is to set IFS to null  See http   pubs opengroup org onlinepubs 009695399 utilities xcu chap02 html tag 02 06 05       If the value of IFS is null  no field splitting shall be performed    Setting to null means setting to empty  value   IFS      Test    ks localhost     echo -n   IFS    cat -vte   I   ks localhost     var   key nvalue   ks localhost     echo  var key value  ks localhost     IFS   ks localhost     echo  var key value  ks localhost

User · Answer

In all of the cases above  the variable is correctly set  but not correctly read  The right way is to use double quotes when referencing   echo   var    This gives the expected value in all the examples given  Always quote variable references     Why    When a variable is unquoted  it will    Undergo field splitting where the value is split into multiple words on whitespace  by default    Before     Foobar is free software     After      Foobar  is  free  software     Each of these words will undergo pathname expansion  where patterns are expanded into matching files   Before      After   bin   boot   dev   etc   home      Finally  all the arguments are passed to echo  which writes them out separated by single spaces  giving   bin  boot  dev  etc  home Foobar is free software Desktop  Downloads    instead of the variable s value    When the variable is quoted it will    Be substituted for its value  There is no step 2    This is why you should always quote all variable references  unless you specifically require word splitting and pathname expansion  Tools like shellcheck are there to help  and will warn about missing quotes in all the cases above

User · Answer

You may want to know why this is happening  Together with the great explanation by that other guy  find a reference of Why does my shell script choke on whitespace or other special characters  written by Gilles in Unix  amp  Linux      Why do I need to write   foo   What happens without the quotes        foo does not mean    take the value of the variable foo     It means   something much more complex          First  take the value of the variable      Field splitting  treat that value as a whitespace-separated list of fields  and build the resulting list  For example  if the variable   contains foo    bar   then the result of this step is the 3-element   list foo     bar      Filename generation  treat each field as a glob  i e  as a wildcard pattern  and replace it by the list of file names that match this   pattern  If the pattern doesn t match any files  it is left   unmodified  In our example  this results in the list containing foo    following by the list of files in the current directory  and finally   bar  If the current directory is empty  the result is foo       bar          Note that the result is a list of strings  There are two contexts in   shell syntax  list context and string context  Field splitting and   filename generation only happen in list context  but that s most of   the time  Double quotes delimit a string context  the whole   double-quoted string is a single string  not to be split   Exception         to expand to the list of positional parameters  e g       is   equivalent to   1    2    3  if there are three positional   parameters  See What is the difference between    and           The same happens to command substitution with   foo  or with     foo   On a side note  don t use  foo   its quoting rules are   weird and non-portable  and all modern shells support   foo  which   is absolutely equivalent except for having intuitive quoting rules       The output of arithmetic substitution also undergoes the same   expansions  but that isn t normally a concern as it only contains   non-expandable characters  assuming IFS doesn t contain digits or   -        See When is double-quoting necessary  for more details about the   cases when you can leave out the quotes       Unless you mean for all this rigmarole to happen  just remember to   always use double quotes around variable and command substitutions  Do   take care  leaving out the quotes can lead not just to errors but to   security   holes

User · Answer

The answer from ks1322 helped me to identify the issue while using docker-compose exec   If you omit the -T flag  docker-compose exec add a special character that break output  we see b instead of 1b     test    usr local bin docker-compose exec db bash -c  echo 1     echo    test b  b   echo    test     cat -vte 1 M    With -T flag  docker-compose exec works as expected     test    usr local bin docker-compose exec -T db bash -c  echo 1     echo    test b  1b

[bash] I just assigned a variable, but echo $variable shows something else

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