c++ - EXCEPTION_CONTINUE_EXECUTION奇怪的行为

我写了代码

void SEHtest(int i) {
  int s = 0;
  __try {
    cout << "code1" << endl;
    int j = 1 / s;
    cout << "code2" << endl;
  } __except((s = 1, i)) {
    cout << "code3" << endl;
  }
  cout << "code4" << endl;
  return;
}
int main() {
  SEHtest(-1);
  return 0;
}

我正在等待输出

code1
code2
code4

但我只有

code1

和无限循环。

为什么？

将volatile密钥名添加到s和j不能解决问题。

最佳答案

导致无限循环是因为每次您恢复执行时都会重新引发异常。不必在过滤器中设置s = 1的值，因为执行是从引起陷阱的指令（在这种情况下为零除）中恢复执行的。如果按照以下方式重新组织代码，您将看到不断抛出异常：

int ExceptionFilter(int& s) {
  cout << "exception filter with s = " << s << endl;
  s++;
  return -1; // EXCEPTION_CONTINUE_EXECUTION
}

void SEHtest() {
  int s = 0;
  __try {
    cout << "before exception" << endl;
    int j = 1 / s;
    cout << "after exception" << endl;
  } __except(ExceptionFilter(s)) {
    cout << "exception handler" << endl;
  }
  cout << "after try-catch" << endl;
  return;
}

int main() {
  SEHtest();
  return 0;
}

结果应为：

before exception
exception filter with s = 0
exception filter with s = 1
exception filter with s = 2
...

继续抛出异常是因为执行是在被零除的指令上执行，而不是在加载s值的指令上执行。这些步骤是：

1  set a register to 0
2  store that register in s (might be optimized out)
3  enter try block
4  output "before exception"
5  load a register from s
6  divide 1 by register (trigger exception)
7  jump to exception filter
8  in filter increment/change s
9  filter returns -1
10 execution continues on line 6 above
6  divide 1 by register (trigger exception)
7  jump to exception filter
8  in filter increment/change s
9  filter returns -1
10 execution continues on line 6 above
...

我认为您无法从该例外中恢复。

关于c++ - EXCEPTION_CONTINUE_EXECUTION奇怪的行为，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/26935823/