Bash: double equals vs -eq

I am doing integer comparison in bash (trying to see if the user is running as root), and I found two different ways of doing it:

Double equals:

if [ $UID == 0 ]
then
fi

-eq

if [ $UID -eq 0 ]
then
fi

I understand that there’s no >= or <= in bash, only -ge and -le, so why is there a == if there’s a -eq?

Is there a difference in the way it compares both sides?

Answers:

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Method 1

== is a bash-specific alias for =, which performs a string (lexical) comparison instead of the -eq numeric comparison. (It’s backwards from Perl: the word-style operators are numeric, the symbolic ones lexical.)

Method 2

To elaborate on bollovan’s answer…

There is no >= or <= comparison operator for strings. But you could use them with the ((...)) arithmetic command to compare integers.

You can also use the other string comparison operators (==, !=, <, >, but not =) to compare integers if you use them inside ((...)).

Examples

• Both [[ 01 -eq 1 ]] and (( 01 == 1 )) do integer comparisons. Both are true.
• Both [[ 01 == 1 ]] and [ 01 = 1 ] do string comparisons. Both are false.
• Both (( 01 -eq 1 )) and (( 01 = 1 )) will return an error.

Note: The double bracket syntax [[...]] and the double parentheses syntax ((...)) are not supported by all shells.

Method 3

If you want to do integer comparison you will better use (( )), where you can also use >= etc.

Example:

if (( $UID == 0 )); then
   echo "You are root"
else
   echo "You are not root"
fi

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0