Why the output of echo and printf is different when I set IFS

Question

I am using Bash 5.0.17 on Ubuntu 20.04

When I run the following commands:

IFS=":"; for i in "1:2:3"; do echo $i; done
# output is: 1 2 3

IFS=":"; for i in "1:2:3"; do echo "$i"; done
output is: 1:2:3

IFS=":"; for i in "1:2:3"; do printf "%s\n" $i; done
# output is: 
# 1
# 2
# 3

IFS=":"; for i in "1:2:3"; do printf "%s\n" "$i"; done
# output is: 1:2:3

This is confusing for me.

1. Why doesn't echo print each token in a separate line?
2. Why does printf works as expected when $i is not quoted?
3. Why both echo and printf fail when $i is quoted?

I appreciate your help

Answer 1

The loop for i in "1:2:3"; do ... runs just once. There's no word splitting of static literal strings, only expansions, and even then only unquoted expansions. You'd see different results with e.g.

IFS=:
var="1:2:3"
for i in $var; do...

The rest is just how echo and printf work. echo joins the arguments with spaces, and prints the joined string followed by a single newline. But printf repeats the format string as many times as necessary to accommodate all arguments, so you get multiple newlines.

So, in the first one, echo $i runs with i set to 1:2:3. The expansion is unquoted, so it's split, and echo get the three arguments 1, 2, 3. It joins them with spaces, giving 1 2 3, and the output is that plus the newline. That's the same as running echo 1 2 3. (Or even echo 1 2 "3", since the number of unquoted spaces on the shell command line doesn't matter.)

The third one is similarly the same as printf "%s\n" 1 2 3, and with the format string repeated, the output is on three lines. Something like printf "%s %s\n" 1 2 3 would use two arguments per repetition, and would print 1 2 on one line, and 3 on another.