Why isn't mkdir -p working right in a script called by checkinstall?

I'm trying to compile Quarter and package it using checkinstall.

If I do the standard ./configure && make && sudo make install, things go fine.

$ wget http://ftp.coin3d.org/coin/src/all/Quarter-1.0.0.tar.gz
$ tar xzf Quarter-1.0.0.tar.gz
$ cd Quarter-1.0.0
$ ./configure
$ make
$ sudo make install

But when I use checkinstall, it fails on a mkdir -p that should work perfectly fine. The way it fails is exactly how it would as if the -p option weren't given. This is the checkinstall command line I'm using:

$ checkinstall -D -y --install=no --pkgname=libquarter --pkgversion=1.0.0 \
  --arch=i386 --pkglicense=GPL --maintainer=me@example.com --reset-uids=yes

This is the failure:

....
/bin/bash ../../../cfg/mkinstalldirs /usr/local/include/Quarter/devices
mkdir -p -- /usr/local/include/Quarter/devices
mkdir: cannot create directory `/usr/local/include/Quarter': No such file or directory
make[4]: *** [install-libdevicesincHEADERS] Error 1
....

This is the relevant part of the script:

$ cat cfg/mkinstalldirs
....
case $dirmode in
  '')
    if mkdir -p -- . 2>/dev/null; then
      echo "mkdir -p -- $*"
      exec mkdir -p -- "$@"
    fi
    ;;
....

I don't understand why that exec is there -- doesn't that guarantee that the remainder of the script (after the esac) will never execute? (If the if test passes, then the script assumes mkdir -p works correctly, so once it does the real mkdir -p it can quit; otherwise the remainder of the script implements proper mkdir -p behavior.) I also don't understand why it uses "$*" in the echo and "$@"<