C - A question of sizeof()

Question

I understand a char[] is "\0" terminated so it has a size of 4, but why the second sizeof() gives 8, since "programmer" has 10 chars?

#include <stdio.h>

int main() {
const char firstname[] = "dev";
const char* lastname = "programmer";
printf("%lu\n", sizeof(firstname));
printf("%lu\n", sizeof(lastname));
return 0;
}

Answer 1

lastname is a pointer variable. lastname does not contain the string "programmer", it has the address of that string. 8 bytes are used to store address value. even if you assign 1000 characters long string to lastname, the size of lastname will remain unchanged - 8 bytes.

Answer 2

I think u might be running the above code in 64 bit system. In 64 bit system, memory address is 64 bits=8 byte.
Also, ur printing the address of a pointer, size of pointer remains same irrespective of data type it points too...
U can know the system size by printing sizeof(int). If answer is 4,then it is 32 bit system. If answer is 8,then it is 64 bit system.