trap in shell programming

[嘉谟]

I want to know  when i use

trap 'rm -f /tmp/my_tmp_file_$$'   INT

why shell itself  capture the single  and she shellscript exit.

Here is the code from << Beginning Programming>>

#!/bin/sh

trap 'rm -f /tmp/my_tmp_file_$$' INT
echo creating file /tmp/my_tmp_file_$$
date > /tmp/my_tmp_file_$$

echo "Press interrupt (Ctrl-C) to interrupt...."
while [ -f /tmp/my_tmp_file_$$ ]; do
    echo File exists
    sleep 1
done
echo The file no longer exists

trap INT
echo creating file /tmp/my_tmp_file_$$
date > /tmp/my_tmp_file_$$

echo "Press interrupt (Ctrl-C) to interrupt...."
while [ -f /tmp/my_tmp_file_$$ ]; do
    echo File exists
    sleep 1
done

echo We never get here

exit 0