Question

我有以下c代码

void *curr = (unsigned long *)some_fixed_address;
if (!curr) {*(unsigned long *)curr = a^b;}

其中some_fixed_address存储长整数0，而a和b是一些预定义的整数。当我运行这段代码时，我得到一个错误

Program received signal SIGSEGV, Segmentation fault.

有什么可能的原因吗？

更新：

这是最小的可验证示例：

#include <stdio.h>

int main()
{
    size_t start = 0;
    size_t a = 5;
    size_t b = 100;
    void *curr = (char *)start;
    if (!curr) {*(unsigned long *)curr = a^b;}
    printf("%d", *(int *)curr);
    return 0;
}

Answer 1

在您的计算机上，NULL可能与0逐位相等（尽管标准不需要这样做）。

因此，当您将some_fixed_address转换为指针时，最终将得到充满0的指针，即NULL指针。

!curr然后检查curr是NULL。然后，您尝试在curr语句的正文中取消引用if，这将导致SEGFAULT。

Answer 2

给出的唯一逻辑解释是@Controller public class EmployeeController { @RequestMapping(value = "/getEmployees", method = RequestMethod.GET) public ModelAndView getEmployeeInfo() { return new ModelAndView("getEmployees"); } @RequestMapping(value = "/showEmployees", method = RequestMethod.GET) public ModelAndView showEmployees(@RequestParam("code") String code) throws JsonProcessingException, IOException { ResponseEntity<String> response = null; System.out.println("Authorization Code------" + code); RestTemplate restTemplate = new RestTemplate(); // According OAuth documentation we need to send the client id and secret key in the header for authentication String credentials = "javainuse:secret"; String encodedCredentials = new String(Base64.encodeBase64(credentials.getBytes())); HttpHeaders headers = new HttpHeaders(); headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON)); headers.add("Authorization", "Basic " + encodedCredentials); HttpEntity<String> request = new HttpEntity<String>(headers); String access_token_url = "http://localhost:8080/oauth/token"; access_token_url += "?code=" + code; access_token_url += "&grant_type=authorization_code"; access_token_url += "&redirect_uri=http://localhost:8090/showEmployees"; response = restTemplate.exchange(access_token_url, HttpMethod.POST, request, String.class); System.out.println("Access Token Response ---------" + response.getBody()); return null; } }为零，这是不可寻址的，因此是段错误。

您的some_fixed_address否则无法执行-仅在if (!curr) { ... }为curr时执行。

我想知道您的0语句的逻辑是否倒退。（或者您没有提供足够的信息，并且您的错误在其他地方。）

Answer 3

这段代码没有意义

void *curr = (unsigned long *)some_fixed_address;
if (!curr) {*(unsigned long *)curr = a^b;} <<< only if curr is null then try to do something with it

我想你是说

void *curr = (unsigned long *)some_fixed_address;
if (curr) {*(unsigned long *)curr = a^b;} <<<< if curr is not null then try to do something

但是，您在该行上遇到错误的事实强烈表明some_fixed_address为0。因此，上面的“更正”代码将运行，但不会执行任何操作。

为什么此C代码会引发分段错误？

3 个答案: