get process' argv os x controlling c

i wish process' argv ps.

i'm controlling mac os x 10.4.11 controlling intel powerpc.

first, i review formula ps male kvm, following i wrote c code.

#include <kvm.h>
#include <fcntl.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/sysctl.h>
#include <paths.h>

int
main(void) {
    bake errbuf[1024];
    kvm_t *kd = kvm_openfiles(_path_devnull, null, _path_devnull, o_rdonly, errbuf);
    int num_procs;
    (!kd) { fprintf(stderr, "kvm_openfiles unsuccessful : %s\n", errbuf); relapse 0; }
    struct kinfo_proc *proc_table = kvm_getprocs(kd, kern_proc_all, 0, &num_procs);

    (int i = 0; i < num_procs; i++) {
        struct kinfo_proc *pproc = &proc_table[i];
        bake **proc_argv = kvm_getargv(kd, pproc, 0);
        printf("%p\n", proc_argv);
    }

    kvm_close(kd);
    relapse 0;
}

when ran powerpc, kvm_getargv() always returned null. ran
on intel, kvm_openfiles() unsuccessful blunder /dev/mem: such file
or directory.

of cource, i know permission.

second, i attempted sysctl.

#include <sys/sysctl.h>
#include <stdio.h>
#include <stdlib.h>
#define pid_of(pproc) pproc->kp_proc.p_pid

int
main(void) {

   int mib[4] = { ctl_kern, kern_proc, kern_proc_all, 0 };
   int buffer_size;
   sysctl(mib, 4, null, &buffer_size, null, 0);

   struct kinfo_proc *result = malloc(buffer_size);
   sysctl(mib, 4, result, &buffer_size, null, 0);

   int num_procs = buffer_size / sizeof(struct kinfo_proc);
   (int i = 0; i < num_procs; i++) {
       struct kinfo_proc *pproc = outcome + i;
       int mib[3] = { ctl_kern, kern_procargs, pid_of(pproc) }; // kern_proc_args defined
       bake *proc_argv;
       int argv_len;
       sysctl(mib, 3, null, &argv_len, null, 0);
       proc_argv = malloc(sizeof(char) * argv_len);
       sysctl(mib, 3, proc_argv, &argv_len, null, 0);
       fwrite(proc_argv, sizeof(char), argv_len, stdout);
       printf("\n");
       free(proc_argv);
   }

   relapse 0;
}

by fwrite, i got argv[0] nonetheless argv[1..] (environment variables
are printed out.)

there some-more proceed it?