Why is shell removing whitespaces when using substring?

Question

When a substring having a space in the beginning is printed the leading whitespace is removed.

$ line="C , D,E,";
$ echo "Output-`echo ${line:3}`";
Output-D,E,

Why is the leading whitespace being removed from the output and how can I print the whitespace?

Answer 1

The substring operation

${line:3}

would extract all characters starting at position 3, which indeed is: [ D,E,] (I added [] just for readability).

However, in command substitution echo ${line:3}, shell performs word splitting and removes the leading blank character, giving the result as D,E,.

Put the substring expression in double quotes to preserve the leading blank, like this:

echo "Output-"`echo "${line:3}"` # => Output- D,E,

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To understand this more clearly, try this:

line="C , D,E,"             # $() does command substitution like backquotes; it is a much better syntax
string1=$(echo "${line:3}") # double quotes prevent word splitting
string2=$(echo ${line:3})   # no quotes, shell does word splitting
string3="$(echo ${line:3})" # since double quotes are outside but not inside $(), shell still does word splitting
echo "string1=[$string1], string2=[$string2], string3=[$string3]" 

gives the output:

string1=[ D,E,], string2=[D,E,], string3=[D,E,]

---

See also:

• Word splitting - GNU documentation
• Word splitting - Greg's Wiki

• Word splitting in Bash with IFS set to a non-whitespace character

• What is the benefit of using $() instead of backticks in shell scripts?

Answer 2

Shell does not preserve multiple spaces between arguments, so the two spaces collapse to one.

echo "Output-`echo ${line:3}`";

Evaluates to:

echo "Output-`echo  -D,E,`";

And:

echo  -D,E,

Evaluates to:

-D,E,

In other words echo receives the same arguments for both of those calls:

echo  -D,E,
echo -D,E,

As the argument arrive as an array of Strings, end each String is already trimmed (no spaces around it).