how many is can you find in:
q: providing command line arguments for an executable through a shell script suppose i have an executable xyz that takes a variable number of command-line arguments, and a wrapper korn shell script xyz.ksh.  is there an easy way to pass all of the shell script arguments as-is to the executable?

a: you need to use:
$@

for correct parameter expansion in all cases. this behaviour is the same in both bash and ksh.
most of the time, $* or $@ will give you what you want. however, they expand parameters with spaces in it. $* gives you all parameters reduced down to one. $@ gives you what was actually passed to the wrapper script.
see for yourself (again, under either bash or ksh):
[tla ~]$ touch file1 file2 space\ file
[tla ~]$ ( test() { ls $*; }; test file1 file2 space\ file )
ls: cannot access space: no such file or directory
ls: cannot access file: no such file or directory
file1  file2
[tla ~]$ ( test() { ls $@; }; test file1 file2 space\ file )
ls: cannot access space: no such file or directory
ls: cannot access file: no such file or directory
file1  file2
[tla ~]$ ( test() { ls $*; }; test file1 file2 space\ file )
ls: cannot access file1 file2 space file: no such file or directory
[tla ~]$ ( test() { ls $@; }; test file1 file2 space\ file )
file1  file2  space file


a: i think youre looking for the $* variable: 
http://www.well.ox.ac.uk/~johnb/comp/unix/ksh.html#commandlineargs
its otherwise known as $@ in bash, i think. 

a: yes.  use the $* variable.  try this script:
#!/bin/ksh

echo $*

and then invoke the script with something like:
scriptname a b c foobar

your output will be:
a b c foobar


user
In the text, there are 87 "i"s.
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