shell $'somestring'

IFS默认支持 空格 tab 换行符

改变 IFS 

IFS=$'\n'

IFS='\n'

IFS=\n

Normally bash doesn't interpret escape sequences in string literals. So if you write \n or "\n" or'\n', that's not a linebreak - it's the letter n (in the first case) or a backslash followed by the letter n(in the other two cases).

$'somestring' is a syntax for string literals with escape sequences. So unlike '\n', $'\n'actually is a linebreak.