Why do I need to quote variable for if, but not for echo?

Question

I've read that you need double quotes for expanding variables, e.g.

if [ -n "$test" ]; then echo '$test ok'; else echo '$test null'; fi

will work as expected, while

if [ -n $test ]; then echo '$test ok'; else echo '$test null'; fi

will always say $test ok even if $test is null.

but then why don't we need quotes in echo $test?

Answer 1

You always need quotes around variables in all list contexts, that is everywhere the variable may be expanded to multiple values unless you do want the 3 side effects of leaving a variable unquoted.

list contexts include arguments to simple commands like [ or echo, the for i in <here>, assignments to arrays... There are other contexts where variables also need to be quoted. Best is to always quote variables unless you've got a very good reason not to.

Think of the absence of quotes (in list contexts) as the split+glob operator.

As if echo $test was echo glob(split("$test")).

The shell behaviour is confusing to most people because in most other languages, you put quotes around fixed strings, like puts("foo"), and not around variables (like puts(var)) while in shell it's the other way round: everything is string in shell, so putting quotes around everything would be cumbersome, you echo test, you don't need to "echo" "test". In shell, quotes are used for something else: prevent some special meaning of some characters and/or affect the behaviour of some expansions.

In [ -n $test ] or echo $test, the shell will split $test (on blanks by default), and then perform filename generation (expand all the *, '?'... patterns to the list of matching files), and then pass that list of arguments to the [ or echo commands.

Again, think of it as "[" "-n" glob(split("$test")) "]". If $test is empty or contains only blanks (spc, tab, nl), then the split+glob operator will return an empty list, so the [ -n $test ] will be "[" "-n" "]", which is a test to check wheter "-n" is the empty string or not. But imagine what would have happened if $test was "*" or "= foo"...

In [ -n "$test" ], [ is passed the four arguments "[", "-n", "" and "]" (without the quotes), which is what we want.

Whether it's echo or [ makes no difference, it's just that echo outputs the same thing whether it's passed an empty argument or no argument at all.

See also this answer to a similar question for more details on the [ command and the [[...]] construct.

Answer 2

@h3rrmiller's answer is good for explaining why you need the quotes for the if (or rather, [/test), but I would actually posit that your question is incorrect.

Try the following commands, and you will see what I mean.

export testvar="123    456"
echo $testvar
echo "$testvar"

Without the quotes, the variable substitution causes the second command to expand to:

echo 123    456

and the multiple spaces are collapsed to a single one:

echo 123 456

With the quotes, the spaces are preserved.

This happens because when you quote a parameter (whether that parameter is passed to echo, test or some other command), the value of that parameter is sent as one value to the command. If you don't quote it, the shell does its normal magic of looking for whitespace to determine where each parameter starts and ends.

This can also be illustrated by the following (very very simple) C program. Try the following on the command line (you may want to do it in an empty directory so as to not risking to overwrite something).

cat <<EOF >paramtest.c
#include <stdio.h>
int main(int argc, char **argv) {
  int nparams = argc-1; /* because 1 parameter means only the executable's name */
  printf("%d parameters received\n", nparams);
  return nparams;
}
EOF
cc -o paramtest paramtest.c

and then...

./paramtest 123 456
./paramtest "123 456"
./paramtest 123   456
./paramtest "123   456"

After running paramtest, $? will hold the number of parameters it was passed (and that number will be printed).

Answer 3

This is all about how the shell interprets the line before a program is executed.

If the line reads echo I am $USER, the shell expands it to echo I am blrfl and echo has no clue whether the origin of the text is a literal or a variable expansion. Similarly, if a line reads echo I am $UNDEFINED, the shell will expand $UNDEFINED into nothing and echo's arguments will be I am, and that's the end of it. Since echo works just fine with no arguments, echo $UNDEFINED is completely valid.

Your issue with if isn't really with if, because if just runs whatever program and arguments follow it and executes the then part if the program exits 0 (or the else part if there is one and the program exits non-0):

if /bin/true ; then echo True dat. ; fi
if fgrep -q blrfl /etc/passwd ; then echo Blrfl has an account. ; fi

When you use if [ ... ] to do a comparison, you're not using primitives built into the shell. You're actually instructing the shell to run a program called [ which is a very slight superset of test(1) that requires its last argument be ]. Both programs exit 0 if the test condition came out true and 1 if it didn't.

The reason some tests break when a variable is undefined is because test doesn't see that you're using a variable. Ergo, [ $UNDEFINED -eq 2 ] breaks because by the time the shell is done with it, all test sees for arguments are -eq 2 ], which isn't a valid test. If you did it with something defined, such as [ $DEFINED -ne 0 ], that would work because the shell would expand it into a valid test (e.g., 0 -ne 0).

There's a semantic difference between foo $UNDEFINED bar, which expands to two arguments (foo and bar) because $UNDEFINED lived up to its name. Compare this with foo "$UNDEFINED" bar, which expands to three arguments (foo, an empty string and `bar). Quotes force the shell to interpret them as an argument whether there's anything between them or not.

Answer 4

Without quotes $test could expand to be more than one word so it needs to be quoted in order to not break the syntax since each switch inside the [ command is expecting one argument which is what the quotes do (makes whatever $test expands to into one argument)

The reason you don't need quotes to expand a variable with echo is because it isn't expecting one argument. It will simply print what you tell it to. So even if $test expands to 100 words echo will still print it.

Take a look at Bash Pitfalls

Answer 5

Empty parameters are removed if not quoted:

start cmd:> strace -e trace=execve echo foo $bar baz
execve("/usr/bin/echo", ["echo", "foo", "baz"], [/* 100 vars */]) = 0

start cmd:> strace -e trace=execve echo foo "$bar" baz
execve("/usr/bin/echo", ["echo", "foo", "", "baz"], [/* 100 vars */]) = 0

The called command doesn't see that there was an empty parameter on the shell command line. It seems that [ is defined to return 0 for -n with nothing following. Whyever.

Quoting does make a difference for echo, too, in several cases:

var='*'
echo $var
echo "$var"

var="foo        bar"
echo $var
echo "$var"