c++ - GoogleMock 是否让我在模拟类中实现析构函数？

Question

我在运行 make 时不断收到这些错误:

debug/main.o: In function `MockMQAdapter<SomeClass>::MockMQAdapter()':
/source/Tests/testsfoo/MockMQAdapter.h:30: undefined reference to `MQAdapter<SomeClass>::~MQAdapter()'
debug/main.o:(.rodata._ZTVN2TW9MQAdapterI6ThingyEE[_ZTVN2TW9MQAdapterI6ThingyEE]+0x10): undefined reference to MQAdapter<SomeClass>::~MQAdapter()'
debug/main.o:(.rodata._ZTVN2TW9MQAdapterI6ThingyEE[_ZTVN2TW9MQAdapterI6ThingyEE]+0x18): undefined reference to `MQAdapter<SomeClass>::~MQAdapter()'
debug/main.o:(.rodata._ZTVN2TW9MQAdapterI6ThingyEE[_ZTVN2TW9MQAdapterI6ThingyEE]+0x20): undefined reference to `MQAdapter<SomeClass>::publish(std::string const&, std::string &message)'

这是我的代码:

#include <gmock/gmock.h>

template<typename S>
class MQAdapter {
public:
    MQAdapter(const std::string host, uint16_t port);
    virtual ~MQAdapter();
    virtual void publish(const std::string queue, std::string message);
};

MQAdapter::MQAdapter(const std::string host, uint16_t port) {}

//Generated by gmock_gen.py
template <typename T0>
class MockMQAdapter : public MQAdapter<T0> {
 public:
  MOCK_METHOD2_T(publish,
      void(std::string, std::string));
};

我非常仔细地遵循了谷歌模拟指南。我不明白这些错误是什么意思。这是我的测试结果:

TEST(MyTest, ExpectCall) {
  MockMQAdapter<SomeClass> *adapter = new MockMQAdapter<SomeClass>("host", 1);
  EXPECT_CALL(*adapter, publish("hi", "hello"));
  adapter->publish("hi", "hello");
}

Answer 1

您已声明 MQAdapter 析构函数，但未定义它。因此，链接器在尝试解决它时会提示。提供一个定义，default 就足够了，即 virtual ~MQAdapter() = default;。

另一方面，构造函数的定义应该是内联的或者用模板参数限定:

template <typename S>
MQAdapter<S>::MQAdapter(const std::string host, uint16_t port) {}

我猜这是因为这只是一个示例，但您没有对任何内容使用 MQAdapter 模板参数，因此它可能是一个常规类。