c++ - goto 和析构函数兼容吗？

Question

此代码导致未定义的行为:

void some_func() {
  goto undefined;
  {
    T x = T();
    undefined:
  }
}

构造函数没有被调用。

但是这段代码呢？ x 的析构函数会被调用吗？我想会的，但我想确定一下。 :)

void some_func() {
  {
    T x = T();
    goto out;
  }
  out:
}

Answer 1

是的，析构函数将按预期调用，就像您因异常提前退出作用域一样。

标准 6.6/2(跳转语句):

On exit from scope (however accomplished), destructors are called for all constructed objects with automatic storage duration that are declared in that scope, in the reverse order of their declaration.