如果我从“goto”跳出一个catch块，我保证异常对象将被释放吗？ [英] If I jump out of a catch-block with "goto", am I guaranteed that the exception-object will be free'ed?

问题描述

我有这样的代码如下

  try {
 doSomething（）; 
} catch（InterruptException）{
 goto rewind_code; 
} 
 
 if（0）{
 rewind_code：
 longjmp（savepoint，1）; 
} 
  

我的问题是，是由C ++运行时免费存储的异常对象当我 goto 出来的catch块？或者是允许缓存它的运行时间，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面的代码，每次采取倒带代码，我不会泄漏内存（因为 longjmp 不会执行清除代码由编译器发送到函数序言之前或之前）。

解决方案

§6.6/ 2：

从范围退出（无论如何），对所有自动存储持续时间的构造对象调用析构函数（12.4）...

至少如我所看到的那样，完成应该包含一个 goto 。

编辑：好的，根据约翰内斯的评论，我们关心的是§15.1/ 4：

当最后一个处理程序被执行为
异常以除throw之外的任何方式退出;临时对象被销毁，执行
可能释放临时对象的内存;

[...]

在处理程序中的异常声明
中声明的对象销毁后立即发生破坏。

I have such code as follows

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

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

My question is, is the exception object that is stored by the C++ runtime free'ed when I goto out of the catch block? Or is the runtime allowed to cache it until the surrounding function exists or something like that? I simply want to ensure that if I execute above code multiple times, each time taking the rewind code, I won't leak memory (because the longjmp won't execute cleanup code emitted by the compiler into or before function prologues).

解决方案

§6.6/2:

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

At least as I'd read it, "however accomplished" should/does include a goto.

Edit: Okay, based on Johannes's comment, what we care about is §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.

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