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使用setrlimit()设置堆栈大小并引发堆栈溢出/segfault

setrlimit segmentation-fault stack linux c

tur1ng · 技术社区 · 14 年前

为什么现在可以在堆栈上分配具有大小的int数组 8kb foo() ?

#include <stdio.h>
#include <sys/resource.h>

void foo(void);

int main() {
 struct rlimit lim = {1024, 1024};

 if (setrlimit(RLIMIT_STACK, &lim) == -1)
  return 1;

 foo();

 return 0;
}

void foo() {
 unsigned ints[2048];

 printf("foo: %u\n", ints[2047]=42);
}

2 回复 | 直到 14 年前

ninjalj 14 年前

立即设置限制,但仅在尝试分配新堆栈或尝试扩展现有堆栈时才进行检查。一个用于RLIMIT_STACK的grep( or a LXR identifier search

显然,堆栈的初始大小是filename+env strings+arg strings加上分配给 setup_arg_pages (2.6.33中20页 1 , 2 ,2.6.34上的128 Kb 3

总而言之:

initial stack size = MIN(size for filename + arg strings + env strings + extra pages, MAX(size for filename + arg strings + env strings, RLIMIT_STACK))

哪里

size for filename + arg strings + env strings <= MAX(ARG_MAX(32 pages), RLIMIT_STACK/4)

exec-shield 补丁(Fedora,Ubuntu,…)有一个额外的EXEC_STACK_BIAS “(再增加2 MB以覆盖随机效应。) ,请参见对新函数的调用 over_stack_limit() acct_stack_growth() ( [Ubuntu1] [Ubuntu2] , [Ubuntu3]

我编辑了原始程序以显示:

#include <stdio.h>
#include <sys/resource.h>

void foo(void);

int main(int argc, char *argv[]) {
        struct rlimit lim = {1, 1};


        if (argc > 1 && argv[1][0] == '-' && argv[1][8]=='l') {
                printf("limiting stack size\n");
                if (setrlimit(RLIMIT_STACK, &lim) == -1) {
                        printf("rlimit failed\n");
                        return 1;
                }
        }

        foo();

        return 0;
}

void foo() {
        unsigned ints[32768];

        printf("foo: %u\n", ints[2047]=42);
}

$./rl
foo: 42
$./rl -l
limiting stack size
Segmentation fault
$

pmg 14 年前

我想 setrlimit 移动“资源指针”,但在您 exec 程序的新副本。

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/resource.h>

void foo(int chk) {
  unsigned ints[2048];
  ints[2047] = 42;
  printf("foo %d: %u\n", chk, ints[2047]);
}

int main(int argc, char **argv) {
  char *newarg[] = { "argv[0]", "one", "two" };
  char *newenv[] = { NULL };
  struct rlimit lim;

  newarg[0] = argv[0];
  getrlimit(RLIMIT_STACK, &lim);
  printf("lim: %d / %d\n", (int)lim.rlim_cur, (int)lim.rlim_max);
  switch (argc) {
    case 1: /* first call from command line */
      lim.rlim_cur = 65536;
      lim.rlim_max = 65536;
      if (setrlimit(RLIMIT_STACK, &lim) == -1) return EXIT_FAILURE;
      newarg[2] = NULL;
      foo(1);
      execve(argv[0], newarg, newenv);
      break;
    case 2: /* second call */
      lim.rlim_cur = 1024;
      lim.rlim_max = 1024;
      if (setrlimit(RLIMIT_STACK, &lim) == -1) return EXIT_FAILURE;
      foo(2);
      execve(argv[0], newarg, newenv);
      break;
    default: /* third call */
      foo(3);
      break;
  }
  return 0;
}

$ ./a.out 
lim: 8388608 / -1
foo 1: 42
lim: 65536 / 65536
foo 2: 42
Killed

为什么在打印限制之前(在调用foo之前)进程会被终止,我不知道。