Я говорю о 32-битной платформе Intel. Ядро Linux версии 2.6.31-14.
#include <stdio.h>
#include <stdlib.h>
int init_global_var = 10; /* Initialized global variable */
int global_var; /* Uninitialized global variable */
static int init_static_var = 20; /* Initialized static variable in global scope */
static int static_var; /* Uninitialized static variable in global scope */
int main(int argc, char **argv, char **envp)
{
static int init_static_local_var = 30; /* Initialized static local variable */
static int static_local_var; /* Uninitialized static local variable */
int init_local_var = 40; /* Initialized local variable */
int local_var; /* Uninitialized local variable */
char *dynamic_var = (char*)malloc(100); /* Dynamic variable */
printf("Address of initialized global variable: %p\n", &init_global_var);
printf("Address of uninitialized global variable: %p\n", &global_var);
printf("Address of initialized static variable in global scope: %p\n", &init_static_var);
printf("Address of uninitialized static variable in global scope: %p\n", &static_var);
printf("Address of initialized static variable in local scope: %p\n", &init_static_local_var);
printf("Address of uninitialized static variable in local scope: %p\n", &static_local_var);
printf("Address of initialized local variable: %p\n", &init_local_var);
printf("Address of uninitialized local variable: %p\n", &local_var);
printf("Address of function (code): %p\n", &main);
printf("Address of dynamic variable: %p\n", dynamic_var);
printf("Address of environment variable: %p\n", &envp[0]);
char* p=0x0;
printf("%s\n",p);
exit(0);
}
Выход:
naman@naman-laptop ~> ./a.out
Address of initialized global variable: 0x804a020
Address of uninitialized global variable: 0x804a03c
Address of initialized static variable in global scope: 0x804a024
Address of uninitialized static variable in global scope: 0x804a034
Address of initialized static variable in local scope: 0x804a028
Address of uninitialized static variable in local scope: 0x804a038
Address of initialized local variable: 0xbfc11cbc
Address of uninitialized local variable: 0xbfc11cb8
Address of function (code): 0x8048484
Address of dynamic variable: 0x8223008
Address of environment variable: 0xbfc11d7c
fish: Job 1, “./a.out” terminated by signal SIGSEGV (Address boundary error)
В приведенном выше коде у меня следующая путаница.
Почему код лежит на 0x8048484, а не где-то близко к началу виртуальной памяти, например, 0x00000400? Насколько я знаю, раскладка должна быть такой:
Недостаточно памяти ........................................ HighMemory
Text Data BSS Heap.....................Stack Env
Итак, текст не должен лежать так глубоко в памяти. Это должно быть близко к нижней памяти, не так ли?