Bash test: what does "=~" do?

Question

#!/bin/bash
INT=-5

if [[ "$INT" =~ ^-?[0-9]+$ ]]; then

echo "INT is an integer."

else

echo "INT is not an integer." >&2

exit 1

fi

What does the leading ~ do in the starting regular expression?

Answer 1

The ~ is actually part of the operator =~, which performs a regular expression match of the string to its left to the extended regular expression on its right.

[[ "string" =~ pattern ]]

Note that the string should be quoted, and the regular expression shouldn't be quoted (unless you want to match literal strings).

A similar operator is used in the Perl programming language and several other general-purpose and domain-specific languages to perform regular expression matching.

The regular expressions understood by bash are the same as those that GNU grep understands with the -E flag, i.e. the extended set of regular expressions.

---

Somewhat off-topic, but good to know:

When matching against a regular expression containing capturing groups, the part of the string captured by each group is available in the BASH_REMATCH array. The zeroth/first entry in this array corresponds to & in the replacement pattern of sed's substitution command (or $& in Perl), which is the bit of the string that matches the pattern, while the entries at index 1 and onwards correspond to \1, \2, etc. in a sed replacement pattern (or $1, $2 etc. in Perl), i.e. the bits matched by each parenthesis.

Example:

string=$( date +%T )

if [[ "$string" =~ ^([0-9][0-9]):([0-9][0-9]):([0-9][0-9])$ ]]; then
  printf 'Got %s, %s and %s\n' \
    "${BASH_REMATCH[1]}" "${BASH_REMATCH[2]}" "${BASH_REMATCH[3]}"
fi

This may output

Got 09, 19 and 14

if the current time happens to be 09:19:14.

The REMATCH bit of the BASH_REMATCH array name comes from "Regular Expression Match", i.e. "RE-Match".

---

In non-bash Bourne-like shells, one may also use expr for limited regular expression matching (using only basic regular expressions).

A small example:

$ string="hello 123 world"
$ expr "$string" : ".*[^0-9]\([0-9][0-9]*\)"
123

Answer 2

You should read the bash man pages, under the [[ expression ]] section.

An additional binary operator, =~, is available, with the same precedence as == and !=. When it is used, the string to the right of the operator is considered an extended regular expression and matched accordingly (as in regex(3)).

Long story short, =~ is an operator, just like == and !=. It has nothing to do with the actual regex in the string to its right.