I tried the following commands

echo $variable

The output is


whereas echo one|name gives an error No command 'name' found. This is reasonable because bash treats | as a pipe and tries to execute command name with one as input.

But why does echo $variable print one|name? After Parameter and Variable expansion, shouldn't it be equivalent to echo one|name?


GNU bash, version 4.3.11(1)-release (i686-pc-linux-gnu)
  • 1
    How would you like it if it were variable='foo|rm -rf /'? Jan 7 '15 at 13:52

No, it shouldn't, because of the way bash operate the command.

When you type echo one|name, bash parse the command, treats | as a pipe token, so | perform pipeline.

When you type echo $variable, because token parsing occur before variable expansion, bash parsing the command into two parts, echo and $variable. After that, it performs variable expansion, expand $variable to one|name. In this case, one|name is a string, | is a part of string and can not be treated as a pipe token (of course, the token recognition phrase was done). The only thing it can be special if IFS variable contains |, bash will use | as delimiter to perform field spliting:

$ variable='one|name'
$ IFS='|'
$ echo $variable
one name
  • Wow! didn't consider the parsing part.
    – nitishch
    Jan 7 '15 at 13:52

The reason is because of the way Bash expands the variable.

Instead of expanding the variable to one|name it expands the variable to "one|name". So here as the value is enclosed within quotes they are treated as string instead of a command.

Below is the strace output of the command which shows how the command gets expanded.

$ variable='one|name'
$ strace echo $variable 
execve("/bin/echo", ["echo", "one|name"], [/* 33 vars */]) = 0
brk(0)                                  = 0x9cc7000
  • Does this happen for all variable expansions?
    – nitishch
    Jan 7 '15 at 13:40
  • 3
    No, it's not the way bash expands variables. It's the way bash parsing the command.
    – cuonglm
    Jan 7 '15 at 13:46
  • But the strace command is interesting. Didn't know it before.
    – nitishch
    Jan 7 '15 at 13:56
  • yes this is the way variables gets expanded in bash. gnu.org/software/bash/manual/html_node/… give you details on how shell expansion happens. Jan 7 '15 at 13:59

The quotes are part of the string you created.

If you did

echo $variable

the result would be as expected. But you are only able to do this in a script (thanks for the helpful comment).

  • variable=one|name itself would give an error
    – nitishch
    Jan 7 '15 at 13:43
  • True, you have to put it in a script, I adjust the answer. Thank you!
    – ddeimeke
    Jan 7 '15 at 13:46

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