Backticks vs braces in Bash

Question

When I went to answer this question, I was going to use the ${} notation, as I've seen so many times on here that it's preferable to backticks.

However, when I tried

joulesFinal=${echo $joules2 \* $cpu | bc} 

I got the message

-bash: ${echo $joules * $cpu | bc}: bad substitution 

but

joulesFinal=`echo $joules2 \* $cpu | bc` 

works fine. So what other changes do I need to make?

Answer 1

The `` is called Command Substitution and is equivalent to $() (parenthesis), while you are using ${} (curly braces).

So all of these expressions are equal and mean "interpret the command placed inside":

joulesFinal=`echo $joules2 \* $cpu | bc` joulesFinal=$(echo $joules2 \* $cpu | bc) #            v                          v #      ( instead of {                   v #                                 ) instead of } 

While ${} expressions are used for variable substitution.

Note, though, that backticks are deprecated, while $() is POSIX compatible, so you should prefer the latter.

---

From man bash:

Command substitution allows the output of a command to replace the command name. There are two forms:

          $(command)    or           `command` 

Also, `` are more difficult to handle, you cannot nest them for example. See comments below and also Why is $(...) preferred over ... (backticks)?.

Answer 2

They behave slightly differently in a specific case:

$ echo "`echo \"test\" `" test  $ echo "$(echo \"test\" )" "test" 

So backticks silently remove the double quotes.