when PHP cgi binary is called from cgi-bin without cgi-redirect, it parses itself (argv[0] of the binary, whatever that happens to be)! i don't think it represents much of a security problem (it still does to some extent, because it reveals path to php and default settings), and no sane person will run the cgi binary without cgi-redirect, but i don't think its the way its supposed to be either..

here is a simple example; this also works with the php binary itself in place of this binary. 
this results in some binary output and the typical phpinfo() page in the middle:
# cat php.c

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

const char *PHP_BINARY="/path/to/php/bin/php";
const char * dummy="<?php phpinfo(); ?>";

int main(int argc, char *argv[]){
  execl(PHP_BINARY,argv[0],0);
  return 1;
};

p.s. btw this simple wrapper (without the phpinfo() part, or course) can be used as a workaround for the vulnerability with cgi-redirect that resulted in the release of 4.3.1 since it removes parameters before exec'ing php itself..

p.p.s. where can i post "feedback"? i can't seem to find it..