What lives above the last accessible address in the stack? - memory

Question

I've asked people before about why the stack doesn't start at 0x7fff...c before, and was told that typically 0x800... onwards is for the kernel, and the cli args and environment variables live at the top of the user's stack which is why it starts below 0x7fff...c. But I recently tried to examine all the strings with the following program
#include <stdio.h>
#include <string.h>
int main(int argc, const char **argv) {
const char *ptr = argv[0];
while (1) {
printf("%p: %s\n", ptr, ptr);
size_t len = strlen(ptr);
ptr = (void *)ptr + len + 1;
}
}
However, after displaying all my environment variables, I see the following (I compiled the program to an executable called ./t):
0x7ffc19f84fa0: <final env variable string>
0x7ffc19f84fee: _=./t
0x7ffc19f84ff4: ./t
0x7ffc19f84ff8:
0x7ffc19f84ff9:
0x7ffc19f84ffa:
0x7ffc19f84ffb:
0x7ffc19f84ffc:
0x7ffc19f84ffd:
0x7ffc19f84ffe:
0x7ffc19f84fff:
So it appears there's one extra empty byte after the null terminator for the ./t string at bytes 0x7ffc19f84ff4..0x7ffc19f84ff7, and after that I segfault so I guess that's the base of the stack. What actually lives in the remaining "empty" space before kernel memory starts?
Edit: I also tried the following:
global _start
extern print_hex, fgets, puts, print, exit
section .text
_start:
pop rdi
mov rcx, 0
_start_loop:
mov rdi, rsp
call print_hex
pop rdi
call puts
jmp _start_loop
mov rdi, 0
call exit
where print_hex is a routine I wrote elsewhere. It seems this is all I can get
0x00007ffcd272de28
./bin/main
0x00007ffcd272de30
abc
0x00007ffcd272de38
def
0x00007ffcd272de40
ghi
0x00007ffcd272de48
make: *** [Makefile:47: run] Segmentation fault
so it seems that even in _start we don't begin at 0x7fff...