When to Use $(…) vs ${…} in Bash: Command vs Parameter Substitution Explained

Background

Linux and other GNU‑based operating systems rely on Bash (the Bourne Again Shell) as a powerful command‑language interpreter, offering both interactive use and scripting capabilities that extend beyond the original Bourne shell.

Command substitution $(...)

The $(command) construct performs command substitution: the enclosed command is executed in a subshell, its standard output is captured, and the result replaces the expression in the surrounding command.

echo "今天是 $(date). LinuxMi.com 又是美好的一天。"

In this example, date runs first, its output is inserted into the echo statement, and the final string is printed.

Parameter expansion ${...}

The ${parameter} syntax handles parameter (variable) expansion. It allows the shell to replace a variable name with its value and supports additional features such as default values, substring extraction, and indirect expansion. animal=lion Attempting to echo $animals prints nothing because the variable animals is undefined. Using parameter expansion correctly yields the desired result: echo ${animal}s This prints lions, demonstrating how braces disambiguate the variable name from following characters.

Differences and precedence

While both forms replace text, they serve distinct purposes: $(...) runs a command and substitutes its output, whereas ${...} substitutes the value of an existing variable. Bash also supports indirect expansion with ${!var}, where the value of var is treated as the name of another variable.

animal=lion
lion=rafiki
echo ${!animal}   # expands to the value of variable named by $animal, i.e., "rafiki"

In such cases, the indirect expansion takes precedence over the direct variable value.

Conclusion

Understanding when to use $(command) for command substitution and ${parameter} for variable expansion helps write clearer, more reliable Bash scripts, avoiding common pitfalls related to quoting, naming, and expansion order.