c++ - throw 在 catch 省略号 (...) 中是否会重新抛出 C++ 中的原始错误？

Question

如果在我的代码中有以下代码段:

try {
  doSomething();
} catch (...) {
  doSomethingElse();
  throw;
}

throw 是否会重新抛出默认省略号处理程序捕获的特定异常？

Answer 1

是的。异常在被捕获之前一直处于事件状态，此时它变为非事件状态。但是它会一直存在到处理程序的范围结束。从标准来看，强调我的:

§15.1/4: The memory for the temporary copy of the exception being thrown is allocated in an unspecified way, except as noted in 3.7.4.1. The temporary persists as long as there is a handler being executed for that exception.

即:

catch(...)
{ // <--

    /* ... */

} // <--

在这些箭头之间，您可以重新抛出异常。只有在处理程序范围结束时，异常才会停止存在。

事实上，在 §15.1/6 中给出的示例与您的代码几乎相同:

try {
    // ...
}
catch (...) { // catch all exceptions
    // respond (partially) to exception <-- ! :D
    throw; //pass the exception to some
           // other handler
}

请记住，如果您 throw 没有事件异常，则将调用 terminate。在处理程序中，这对您来说不是这种情况。

---

如果 doSomethingElse() 抛出并且异常没有相应的处理程序，因为原始异常被认为已处理，新异常将替换它。 (好像它刚刚抛出，开始堆栈展开等)

即:

void doSomethingElse(void)
{
    try
    {
        throw "this is fine";
    }
    catch(...)
    {
        // the previous exception dies, back to
        // using the original exception
    }

    try
    {
        // rethrow the exception that was
        // active when doSomethingElse was called
        throw; 
    }
    catch (...)
    {
        throw; // and let it go again
    }

    throw "this replaces the old exception";
    // this new one takes over, begins stack unwinding
    // leaves the catch's scope, old exception is done living,
    // and now back to normal exception stuff
}

try
{
    throw "original exception";
}
catch (...)
{
  doSomethingElse();
  throw; // this won't actually be reached,
         // the new exception has begun propagating
}

当然，如果没有抛出任何异常，throw; 将被触发，您将按预期抛出捕获的异常。