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POGet other process' argv in OS X using C
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copied!<p>I want to get other process' argv like ps.</p> <p>I'm using Mac OS X 10.4.11 running on Intel or PowerPC.</p> <p>First, I read code of ps and man kvm, then I wrote some C code.</p> <pre><code>#include <kvm.h> #include <fcntl.h> #include <stdio.h> #include <stdlib.h> #include <sys/sysctl.h> #include <paths.h> int main(void) { char errbuf[1024]; kvm_t *kd = kvm_openfiles(_PATH_DEVNULL, NULL, _PATH_DEVNULL, O_RDONLY, errbuf); int num_procs; if (!kd) { fprintf(stderr, "kvm_openfiles failed : %s\n", errbuf); return 0; } struct kinfo_proc *proc_table = kvm_getprocs(kd, KERN_PROC_ALL, 0, &num_procs); for (int i = 0; i < num_procs; i++) { struct kinfo_proc *pproc = &proc_table[i]; char **proc_argv = kvm_getargv(kd, pproc, 0); printf("%p\n", proc_argv); } kvm_close(kd); return 0; } </code></pre> <p>When ran on PowerPC, <code>kvm_getargv()</code> always returned NULL. When ran on Intel, <code>kvm_openfiles()</code> failed with error <code>/dev/mem: No such file or directory</code>.</p> <p>Of cource, I know about permission.</p> <p>Second, I tried sysctl.</p> <pre><code>#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); for (int i = 0; i < num_procs; i++) { struct kinfo_proc *pproc = result + i; int mib[3] = { CTL_KERN, KERN_PROCARGS, pid_of(pproc) }; // KERN_PROC_ARGS is not defined char *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); } return 0; } </code></pre> <p>By fwrite, I got argv[0] but argv[1..] are not (environment variables are printed out.)</p> <p>There is no more way to do it?</p>

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