c++ - 如果我跳出带有 "goto"的捕获 block ，我是否保证异常对象将被释放？

Question

我有如下代码

try {
  doSomething();
} catch(InterruptException) {
  goto rewind_code;
}

if(0) {
rewind_code:
  longjmp(savepoint, 1);
}

我的问题是，当我 goto 离开 catch block 时，C++ 运行时存储的异常对象是否被释放？或者运行时是否允许缓存它，直到周围的函数存在或类似的东西？我只是想确保如果我多次执行上述代码，每次使用倒带代码时，我都不会泄漏内存(因为 longjmp 不会执行编译器发出的清理代码到或在函数序言之前)。

Answer 1

§6.6/2:

On exit from a scope (however accomplished), destructors (12.4) are called for all constructed objects with automatic storage duration...

至少正如我所读到的那样，“无论如何完成”应该/确实包含一个 goto。

编辑:好的，根据 Johannes 的评论，我们关心的是 §15.1/4:

When the last handler being executed for the exception exits by any means other than throw; the temporary object is destroyed and the implementation may deallocate the memory for the temporary object;

[...]

The destruction occurs immediately after the destruction of the object declared in the exception-declaration in the handler.