What is the difference between "$a" and $a in unix

Question

For example:

#!/bin/sh
a=0
while [ "$a" -lt 10 ]
   b="$a"
   while [ "$b" -ge 0 ] do
      echo -n "$b "
     b=`expr $b - 1`
   done
   echo
   a=`expr $a + 1`
done*

The above mentioned script gives the answer in triangle while with out the double quotes, it falls one after the other on diff lines.

Answer 1

After a variable is expanded to its value, word splitting (i.e. separating the value into tokens at whitespace) and filename wildcard expansion takes place unless the variable is inside double quotes.

Example:

var='foo   bar'
echo No quotes: $var
echo With quotes: "$var"

will output:

No quotes: foo bar
With quotes: foo   bar

Answer 2

Here the difference is how the argument is passed to echo function. Effectively " " will preserve whitespaces.

This:

echo -n "$b "

Is translated to:

echo -n "<number><space>"

While this:

echo -n $b<space>

Will ignore the trailing space and will just output the number:

echo -n <number>

Therefore removing all the spaces that are needed for output to look "triangular".

Answer 3

There are errors in your script:

• no do after 1st while
• no ; before do after 2nd while
• why asterisk on done* at the end?

Now to answer your question. If used as a paramenter:

• "$a" is one argument.
• $a (without quotes) is possibly multiple arguments:

Compare:

v='a b';  set $v; echo "\$#=$#, \$1=\"$1\",  \$2=\"$2\""  
$#=2, $1="a",  $2="b"

v='a b'; set "$v"; echo "\$#=$#, \$1=\"$1\",  \$2=\"$2\""  
$#=1, $1="a b",  $2=""