关于Linux进程内存布局的问题
作者:互联网
我说的是英特尔32位平台. Linux内核版本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);
}
输出:
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)
在上面的代码中,我有以下困惑.
为什么代码位于0x8048484而不是虚拟内存开头附近,比如0x00000400?据我所知,布局应该是这样的:
内存不足…………………………………. HighMemory
Text Data BSS Heap.....................Stack Env
所以,文字不应该落在记忆中.它应该接近较低的记忆,不应该吗?
解决方法:
Why does the code lie at 0x8048484
因为默认加载地址(将在此地址加载ELF文件的开头)是0x8000000(或0x8048000).此缺省值在缺省链接器(ld)脚本中得到修复,可以通过链接器选项进行更改.
注意,这是0x
Why does the code lie at 0x8048484
000或0x08048000(128兆字节)而不是0x80000000(2千兆字节).
以下是关于此论坛的限制
Why does the code lie at 0x8048484
标签:linux,process,memory-layout 来源: https://codeday.me/bug/20190621/1254075.html