相关疑难解决方法(0)

知道为什么信号处理程序会进入无限循环吗？

这是代码.请帮我.

enter code here
 9 void SIGSEGV_handler(int signal)
10 {
11  printf("Segmentation fault caught....\n");
12  printf("Value of instance variable: i = %d\n\n", i);
13 } 
16 
17 int main()
18 {
19  char *mallocPtr, *callocPtr, *reallocPtr, *memalignPtr, *vallocPtr;
20  struct sigaction sa;
21 
22  sa.sa_handler=SIGSEGV_handler;
23  sigaction(SIGSEGV, &sa, NULL);
24 
37 
38  printf("The segmentation fault handler will be entered for i = 3, 4, 5 and 6\n");
39 
40 
41  for(i=0; i<7; i++)
42   {
43    printf("i = %d\n",i); …

我使用sigsetjmp和siglongjmp编写了一个分段错误处理程序.一旦它进入信号处理程序,我调用siglongjmp,以便跳过错误的指令.

问题是,我再次想要导致SIGSEGV并转到相同的处理程序,但现在sigsetjmp将返回1.

如何重置sigsetjmp？

这是我的代码:

#include <stdio.h>
#include <memory.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
#include <sys/types.h>
#include <fcntl.h>
#include <signal.h>
#include <setjmp.h>

sigjmp_buf env, env1;

void SIGSEGV_handler(int signal)
{
  printf("Segmentation fault caught\n");
  siglongjmp(env, 1);
}

int main()
{
 void * allocation;
 size_t size;
 static int devZerofd = -1;
 struct sigaction sa, sa1;

 sa.sa_handler=(void*)SIGSEGV_handler;
 sigaction(SIGSEGV, &sa, NULL);

 if ( devZerofd == -1 ) {
     devZerofd = open("/dev/zero", O_RDWR);
     if ( devZerofd < 0 )
         perror("open() on /dev/zero failed");
  }

 allocation = …