c++ - 重新抛出异常

Question

为什么以下不处理被重新抛出的异常？我尝试了所有组合，但没有一个会显示最后一次捕获的输出，所以我很困惑!

Derived D;

try {
       throw D;
} catch ( const Derived &d) {
       throw;
} catch (const Base &b) {
      cout << "caught!" << endl;
}

---

Derived D;

try {
    throw D;
} catch ( const Derived d) {
    throw;
} catch (const Base b) {
    cout << "caught!" << endl;
}

---

Derived D;

try {
    throw D;
} catch ( const Derived d) {
    throw;
} catch (const Base &b) {
    cout << "caught!" << endl;
}

---

Derived D;

try {
    throw D;
} catch ( const Derived &d) {
    throw;
} catch (const Base b) {
    cout << "caught!" << endl;
}

Answer 1

重新抛出不是由相同的 try-catch block 处理。它被抛出到调用范围。

在 [except.throw](2003 年措辞)中:

A throw-expression with no operand rethrows the exception being handled.

和:

When an exception is thrown, control is transferred to the nearest handler with a matching type (15.3); “nearest” means the handler for which the compound-statement, ctor-initializer, or function-body following the try keyword was most recently entered by the thread of control and not yet exited.

你的 try block 已经退出，所以它的处理程序不是候选的。因此，代码中的任何 catch block 都不能处理重新抛出。

诚然，这是相当令人困惑的措辞。