c++ - 具有多重继承的 shared_ptr 转换

Question

我有点糊涂了。

namespace Io
{
    class IDevice;
}

//...

namespace Sensor
{
    class IDevice;
}

//...

class ComplexDeviceHandler : public Io::IDevice, public Sensor::IDevice;

//...

std::vector<std::shared_ptr<Io::IDevice>> devices; //populated by objects of type ComplexDeviceHandler

//..

for (const auto& device : devices)
{
    std::shared_ptr<Sensor::IDevice> sensor = device; //obviously, error here
}

两者都是 Io::IDevice和 Sensor::IDevice是接口(interface)(某种程度上)。我应该使用什么转换来转换 std::shared_ptr<Io::IDevice>至 std::shared_ptr<Sensor::IDevice> .在这种情况下，std::shared_ptr<Io::IDevice>存储 ComplexDeviceHandler 类型对象的地址，它是两种类型的 child 。

Answer 1

您需要先尝试将其转换为 ComplexDeviceHandler*，检查它是否有效，然后将其转换为 Sensor::IDevice*:

for (const auto& device: devices)
{
    if (auto sidptr = std::dynamic_pointer_cast<ComplexDeviceHandler>(device)) {
        std::shared_ptr<Sensor::IDevice> sensor = sidptr;
    }
}