C++ memcpy of char arrays

I was exploring memcpy in C++. I have a program that reads 10 bytes from a file called temp.txt. The content of the file is:

abcdefghijklmnopqrstuvwxyz

Here's the code:

// This file is called "temp.cpp"

#include <string>
#include <fcntl.h>
#include <iostream>
#include <unistd.h>

int main() {
    int fd = open("temp.txt", O_RDONLY);
    int buffer_size{10};
    char buffer[11];
    char copy_buffer[11];
    std::size_t bytes_read = read(fd, buffer, buffer_size);
    std::cout << "Buffer: " << buffer << std::endl;
    printf("Buffer address: %p, Copy Buffer address: %p\n", &buffer, &copy_buffer);
    memcpy(&copy_buffer, &buffer, 7);
    std::cout << "Copy Buffer: " << copy_buffer << std::endl;
    return 0;
}

I read 10 bytes and store them (and \0 in buffer). I then want to copy the contents of buffer into copy_buffer. I was changing the number of bytes I want to copy in the memcpy() function. Here's the output:

memcpy(&copy_buffer, &buffer, 5); // abcde
memcpy(&copy_buffer, &buffer, 6); // abcdef
memcpy(&copy_buffer, &buffer, 7); // abcdefg
memcpy(&copy_buffer, &buffer, 8); // abcdefgh?C??abcdefghij

I noticed that the last output is weird. I tried printing the addresses of copy_buffer and buffer and here's what I got:

Buffer address: 0x16cf8f5dd, Copy Buffer address: 0x16cf8f5d0

Which means, when I copied 8 characters, copy_buffer did not terminate with a \0, so the cout went over to the next addresses until it found a \0. This explains the entire buffer getting printed since it has a \0 at its end.

Why doesn't the same happen when I memcpy() 5, 6, 7 bytes? Is it because there's a \0 at address 0x16cf8f5d7 which gets overwritten only when I copy 8 bytes?