IFS usage in bash

I have

$ IFS=123

$ myvar=abc1def1ghi

$ for i in "$myvar";  do echo i $i;  done
i abc def ghi

$ for i in $myvar;  do echo i $i;  done
i abc
i def
i ghi

$ myvar="abc1def1ghi"

$ for i in "$myvar";  do echo i $i;  done
i abc def ghi

$ for i in $myvar;  do echo i $i;  done
i abc
i def
i ghi

I think I understand whats happening in 2nd and 4th for loops. But I do not understand why 1 was not printed in 1st and 3rd for loops. The whole idea of IFS is confusing to me in general.

Answer

When you say $var_name then 1 is interpreted as a separator and hence you see the string as cut into pieces.

$ IFS=123

$ myvar=abc1def1ghi

$ echo $myvar
abc def ghi

When you add quotes around the variable, you demand for the absolute value of the variable without any processing.

$ echo "$myvar"
abc1def1ghi

Now coming to your question about 1st loop,

$ for i in "$myvar";  do echo i $i;  done

is equivalent to

$ for i in "abc1def1gh1";  do echo i $i;  done

‘i’ gets assigned “abc1def1gh1” and it prints out the value of variable ‘i’. This time ‘1’ is interpreted as a separator while printing ‘i’. This is what happened while running the loop:

$ i="abc1def1gh1"

$ echo $i
abc def gh

Same thing happens in the 3rd loop.

Now if you want ‘1’ to be printed, then add quotes around $i in the loop. ie. change this:

$ for i in "abc1def1gh1";  do echo i $i;  done
i abc def gh

to

$ for i in "abc1def1gh1";  do echo i "$i";  done
i abc1def1gh1