CODEWITHSUNDEEP
bashshellescaping
Pandav Patel
Pandav Patel

Reputation: 82

Shell scripting: How does escape sequence characters behave while retrieving value of variable with escape sequence in it?

I have test.sh as shown below. When I run it like ./test.sh from bash shell, I get output as shown below.

test.sh

operator="\*"
echo $operator
expr 5 $operator 1
expr 5 \* 2

output

\*
expr: syntax error
10

I am not understanding why does it give syntax error for the third statement while fourth statement is working as expected. If expr is only getting * as second argument in third statement, how does echo output \*?

Upvotes: 1

Views: 97

Answers (1)

muru
muru

Reputation: 4896

The final stage of shell expansions in bash is quote removal:

After the preceding expansions, all unquoted occurrences of the characters ‘\’, ‘'’, and ‘"’ that did not result from one of the above expansions are removed.

The "above expansions" here being parameter (variable) expansion, command substitution, etc.

In expr 5 \* 2, the \ will be removed by bash as part of quote removal, since it is not a result of variable (or other) expansions. So here, the argument that expr gets is *, after the removal of \.

In expr 5 $operator 1 and echo $operator, the \ will not be removed, since it's a result of variable expansion. So in both of these commands, the argument that echo and expr get is \*, without the \ being removed.

The best way here is to use quotes consistently:

operator="*"

echo "$operator"
expr 5 "$operator" 2

Otherwise, \* might still be subject to field splitting, if IFS happened to be set to something weird:

bash-4.4$ foo='\*'
bash-4.4$ echo $foo
\*
bash-4.4$ IFS='*'
bash-4.4$ echo $foo
\
bash-4.4$ echo "$foo"
\*

Upvotes: 1

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