In kernel 3.8, how the first user process is switched into user mode while kernel_execve is removed

Question

In kernel 3.8.x and later version, the definition for run_init_process is changed.

The following is the new definition for run_init_proces in kernel 3.8.

 static int run_init_process(const char *init_filename) {
         argv_init[0] = init_filename;
        return do_execve(init_filename,
                (const char __user *const __user *)argv_init,
                 (const char __user *const __user *)envp_init); }

Compared to the definition in kernel 3.7.x and old version.

static int run_init_process(const char *init_filename) {
         argv_init[0] = init_filename;
         return kernel_execve(init_filename, argv_init, envp_init); }

The most critical part in kernel_execve is that it will call the ret_from_kernel_execve, which will switch into the user mode then.

In the new definition, kernel_execve is gone. My question is how the first user process is switched to the user mode then.

Answer 1

The successful do_execv() sets up the current process to run the new program (e.g. via load_elf_binary()), and then returns 0 to run_init_process(), which returns 0 to kernel_init(), which also returns 0, and was called as part of:

    kernel_thread(kernel_init, NULL, CLONE_FS | CLONE_SIGHAND);

This is where the rules from https://lwn.net/Articles/520227/ come in: our fn() has returned 0 after an execve, so "the thread will proceed into userland context created by that execve".