Question

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，例如0x00000400？据我所知，布局应该是这样的：

内存不足........................................ HighMemory < / strong>

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？

Answer 1

为什么代码位于0x8048484

因为默认加载地址（将在此地址加载ELF文件的开头）为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0x08000000或0x08048000（128兆字节）而不是0x80000000（2千兆字节）。

以下是关于此限制http://cboard.cprogramming.com/tech-board/101129-why-address-space-0-0x08000000-process-unused.html在论坛和http://books.google.com/books?id=Id9cYsIdjIwC&pg=PA111&lpg=PA111&dq=linker+0x08000000书籍中的讨论。同样http://lkml.org/lkml/2002/2/20/194在lkml，描述良好：

“0x8048000是文本段的典型起点符合System V Intel 386 ABI规范（http://stage.caldera.com/developer/devspecs/abi386-4.pdf）。“

Answer 2

什么时候a.out不是a.out？当它实际上是一个ELF。请尝试elfinfo --all a.out了解详情。

关于Linux进程内存布局的问题

2 个答案: