z flag in parameter expansion doesn't use $IFS ?

[Han Pingtian <hanpt@xxxxxxxxxxxxxxxxxx>]

Hello, 

It looks like the z flag of parameter expansion doesn't care what the
value of IFS is:

% zsh -c 'var="foo ''bar baz''";print -l ${(z)var}'
foo
'bar baz'
% zsh -c 'IFS=:;var="foo:''bar:baz''";print -l ${=var}'
foo
'bar
baz'
% zsh -c 'IFS=:;var="foo:''bar:baz''";print -l ${(z)var}'
foo:'bar:baz'
%

I believe z flag should regard IFS and split expansion results just like
the shell parsing. Or I'm missing something here? Please help.

Thanks in advance.