Why can you use -eq in Bash conditionals without adding a '$' before a variable name?

Question

I was running some Bash conditional scripts and discovered if I run this:

#!/bin/bash

read foo

if [[ foo -eq 1 ]]; then
  echo "A"
fi

if [[ foo -eq 2 ]]; then
  echo "B"
fi

The conditionals work fine under Bash 4.2.25 without the use of $foo. Why does this work without referencing the variable with a $?

Basically because it is a bash built-in command (compared with either the Bourne shell or the POSIX shell) and it has been decreed that variables do not need to be prefixed with $ in this one context. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 27, 2013 at 4:27

Barmar · Accepted Answer · 2013-12-27 04:38:54Z

3

From the description of bash Conditional Constructs, it says that [[ expression ]] performs arithmetic expansion of the expression. If you then find the section on Shell Arithmetic it says:

Within an expression, shell variables may also be referenced by name without using the parameter expansion syntax.

"parameter expansion syntax" refers to putting a $ before the name.

answered Dec 27, 2013 at 4:38

Barmar

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