Why does the Bourne shell printf iterate over a %s argument?

Question

What's going on here?

printf.sh:

#! /bin/sh
NAME="George W. Bush"
printf "Hello, %s\n" $NAME

Command line session:

$ ./printf.sh
Hello, George
Hello, W.
Hello, Bush

UPDATE: printf "Hello, %s\n" "$NAME" works. For why I'm not using echo, consider

echo.sh:

#! /bin/sh
FILE="C:\tmp"
echo "Filename: $FILE"

Command-line:

$ ./echo.sh
Filename: C:    mp

The POSIX spec for echo says, "New applications are encouraged to use printf instead of echo" (for this and other reasons).

Answer 1

Your NAME variable is being substituted like this:

printf "Hello, %s\n" George W. Bush

Use this:

#! /bin/sh
NAME="George W. Bush"
printf "Hello, %s\n" "$NAME"

Answer 2

is there a specific reason you are using printf or would echo work for you as well?

NAME="George W. Bush"
echo "Hello, "$NAME

results in

Hello, George W. Bush

edit: The reason it is iterating over "George W. Bush" is because the bourne shell is space delimitted. To keep using printf you have to put $NAME in double quotes

printf "Hello, %s\n" "$NAME"