variable assignment and IFS

Question

Per Greg's Wiki the IFS variable is used:

• In the read command, if multiple variable-name arguments are specified, IFS is used to split the line of input so that each variable gets a single field of the input.
• When performing WordSplitting on an unquoted expansion, IFS is used to split the value of the expansion into multiple words.
• When performing the "$*" or "${array[*]}" expansion,the first character of IFS is placed between the elements in order to construct the final output string.
• When doing "${!prefix*}", the first character of IFS is placed between the variable names to make the output string.
• IFS is used by complete -W under programmable completion

So my question is, why should IFS have to come into play in variable assignment? Per the below, bash is applying word-splitting on the string on the right(a:b:c:d).

$ IFS=: s=a:b:c:d $ echo $s a b c d 

Answer 1

It doesn't. You need to quote properly. Word splitting is applied to unquoted expansions according to the value of IFS. The problem is your echo command, not the assignment.

 $ ( IFS=: s=a:b:c:d typeset -p s ) declare -- s="a:b:c:d"