Ptrace get execve filename

Question

I trying to make a program like strace, but I'm stuck with the syscall execve, the registers rdi, rsi and rdx all contain 0 (I also tried with an assembler file that makes the hard-coded syscall), here's part of the code:

regs = Registers()  # Registers is a ctypes version of user_regs_struct and works well with others syscalls
libc.ptrace(PtraceReq.PTRACE_GETREGS, pid, None, byref(regs))
print(regs.rdi, regs.rsi, regs.rdx)
# 0, 0, 0

How to get the pointer to the file name ?

EDIT: more complete code:

class Tracer:
    def __init__(self):
        pass

    def trace_exec(self, command, *args):
        child_pid = os.fork()
        if child_pid == 0:
            # PtraceReq is PTRACE_* enum
            libc.ptrace(PtraceReq.PTRACE_TRACEME, 0, None, None)
            exec(command, *args)  # choose the best version of exec familly for passed args
        else:
            self._trace(child_pid)

    def _trace(self, pid):
        status = os.wait()[1]
        libc.ptrace(PtraceReq.PTRACE_SETOPTIONS, pid, 0, 1 << 20)  # PTRACE_O_EXITKILL
        while status == 1407:
            libc.ptrace(PtraceReq.PTRACE_SYSCALL, pid, None, None)
            status = os.wait()[1]
            self._get_syscall(pid)  # before this was above os.wait and ptrace_syscall

    def _get_syscall(self, pid):
        regs = Registers()

        # Enter syscall
        libc.ptrace(PtraceReq.PTRACE_GETREGS, pid, None, byref(regs))
        # list of all syscalls by ID, call them parse the arguments
        syscall = syscalls[regs.orig_rax](regs.rdi, regs.rsi, regs.rdx, regs.r10, regs.r8, regs.r9)

        libc.ptrace(PtraceReq.PTRACE_SYSCALL, pid, None, None)
        os.wait()

        # Exit syscall
        libc.ptrace(PtraceReq.PTRACE_GETREGS, pid, None, byref(regs))
        syscall.returned(regs.rax)
        #if (syscall.name == 'connect') or (syscall.name == 'execve'):
        print(syscall.as_function())

if __name__ == '__main__':
    Tracer().trace_exec(sys.argv[1], *sys.argv[2:])