Bash word splitting mechanism

Question

I am new to Bash and I am seeing that there is automatic word splitting done by Bash:

a="1  2     3 4"

If I echo "a" by echo $a I got 1 2 3 4, which has done word splitting implicitly. If I loop through "a", I got 1, 2, 3, and 4 respectively.

I also read from here that

The shell scans the results of parameter expansion, command substitution, and arithmetic expansion that did not occur within double quotes for word splitting.

And I also found that if I have

b=$a;
echo "$b"

I would get

"1 2 3 4"

So, here is my problem: when is the word splitting done? does it change the string itself? Does it only take effect when I use echo or for (loop)?

More generally, how does bash handle it?

Answer 1

There are actually several rounds of word-splitting. The first is performed prior to parsing the command line, so echo $a is split into two words echo and $a. (This is why something like a="echo foo | wc -l"; $a doesn't execute a pipeline; parsing is complete before $a is expanded). After that round of word-splitting is over, parameter expansion occurs to produce 2 strings, echo and 1 2 3 4. The string resulting from parameter expansion then undergoes word-splitting itself, since it is not quoted, producing 4 additional words 1, 2, 3, and 4.

In a for loop, the items in the list are subject to word-splitting:

for b in $a; do

is expanded (after word-splitting produces for, b, in, $a, ;, and do) to for, b, in, 1 2 3 4, ;, and do. Again the string resulting from parameter expansion undergoes word-splitting to 1, 2, 3, and 4.