c++ - 通过引用捕获的抛出对象的生命周期

Question

C++ 标准第 15.1.4 段说:

The memory for the temporary copy of the exception being thrown is allocated in an unspecified way, except as noted in 3.7.3.1. The temporary persists as long as there is a handler being executed for that exception.

我想知道为什么这段代码会崩溃(我知道这不是最佳实践):

class magicException
{
private:
    char* m_message;

public:
    magicException(const char* message)
    {
        m_message = new char[strlen(message) + 1];
        strcpy(m_message, message);
    }

    ~magicException()
    {
        cout << "Destructor called." << endl;
        delete[] m_message;
    }

    char* getMessage()
    {
        return m_message;
    }
};

void someFunction()
{
    throw magicException("Bang!");
}

int main(int argc, char * argv[])
{
    try
    {
        someFunction();
    }
    catch (magicException& ex)
    {
        cout << ex.getMessage() << endl;
    }

    return 0;
}

具体来说，抛出的 magicException 对象的析构函数在 catch block 之前被调用。但是，如果我向我的类添加一个复制构造函数:

magicException(const magicException& other)
{
    cout << "Copy constructor called." << endl;
    m_message = new char[strlen(other.m_message) + 1];
    strcpy(m_message, other.m_message);
}

然后代码开始运行，析构函数在预期的位置(catch block 的末尾)被调用，但有趣的是，复制构造函数仍然没有被调用。它是否被编译器优化掉了(关闭优化的 Visual C++ 2008)，还是我遗漏了什么？

Answer 1

Specifically, the destructor of the thrown magicException object gets called before the catch block.

是的，正如您从标准中引用的那样，编译器会获取一份拷贝，而原始文件(可能)会被丢弃。您的问题是原始代码中缺少复制构造函数。但是，允许 C++ 编译器在各种情况下删除(或添加)复制构造函数调用，包括这种情况。