保护内存不被改变
有没有办法保护内存区域？
我有这个结构：
#define BUFFER 4
struct
{
char s[BUFFER-1];
const char zc;
} str = {'\0'};
printf("'%s', zc=%d\n", str.s, str.zc);
它应该操作长度为 BUFFER-1 的字符串，并保证它以 '\0' 结尾。
下给出错误
str.zc='e'; /*error */
但编译器仅在以下情况
str.s[3]='e'; /*no error */
：如果使用 gcc 和某些标志进行编译可能会出现错误，那也很好。
谢谢， 贝科
Is there a way to protect an area of the memory?
I have this struct:
#define BUFFER 4
struct
{
char s[BUFFER-1];
const char zc;
} str = {'\0'};
printf("'%s', zc=%d\n", str.s, str.zc);
It is supposed to operate strings of lenght BUFFER-1, and garantee that it ends in '\0'.
But compiler gives error only for:
str.zc='e'; /*error */
Not if:
str.s[3]='e'; /*no error */
If compiling with gcc and some flag might do, that is good as well.
Thanks,
Beco
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
要在运行时检测错误，请查看 gcc 中的 -fstack-protector-all 选项。当尝试检测像您所描述的那样的非常小的溢出时，它的用途可能有限。
不幸的是，您不会找到大量有关检测缓冲区溢出情况（如您在编译时描述的情况）的信息。从 C 语言的角度来看，语法是完全正确的，并且该语言为您提供了足够的绳索来吊死自己。如果您确实想保护您的缓冲区免受您自己的影响，您可以编写一个数组访问的前端，在允许访问您想要的内存之前验证索引。
To detect errors at runtime take a look at the -fstack-protector-all option in gcc. It may be of limited use when attempting to detect very small overflows like the one your described.
Unfortunately you aren't going to find a lot of info on detecting buffer overflow scenarios like the one you described at compile-time. From a C language perspective the syntax is totally correct, and the language gives you just enough rope to hang yourself with. If you really want to protect your buffers from yourself you can write a front-end to array accesses that validates the index before it allows access to the memory you want.