3

我有一个 Visual Studio 2008 C++03 项目，我想在其中对一个使用公开静态方法（基于策略的设计、策略模式）的特征模板参数的类进行单元测试。我正在使用 Google Test 和 Google Mock 框架。

例如：

/// the class under test
template< typename FooTraits >
class Foo
{
public:
    void DoSomething()
    {
        FooTraits::handle_type h = FooTraits::Open( "Foo" );
        /* ... */
        FooTraits::Close( h );
    };
};

/// a typical traits structure
struct SomeTraits
{
    typedef HANDLE handle_type;
    static handle_type Open( const char* name ) { /* ... */ };
    static void Close( handle_type h ) { /* ... */ };
};

/// mocked traits that I would like to use for testing
struct MockTraits
{
    typedef int handle_type;
    static MOCK_METHOD1( Open, handle_type( const char* ) );
    static MOCK_METHOD1( Close, void( handle_type ) );
};

/// the test function
TEST( FooTest, VerifyDoSomethingWorks )
{
    Foo< MockTraits > foo_under_test;

    // expect MockTraits::Open is called once
    // expect MockTraits::Close is called once with the parameter returned from Open
    foo_under_test.DoSomething();
};

显然这不会按原样工作。Google Mock 不能模拟静态方法，我需要在测试中创建一个 Mocked 类的实例来设置它的行为和期望。

那么使用 Google Test / Google Mock 对接受模板策略的类进行单元测试的正确方法是什么？

4

1 回答 1

3

你能用非静态方法创建一个类，创建它的全局实例（或在你的特征中是静态的），并让你的特征类服从它吗？

因此，为了澄清受 Rob 评论启发的想法：

struct FunnyDoodad
{
   FunnyDoodad();
   ~FunnyDoodad();

   MOCK_METHOD1( Open, HANDLE( const char* ) );
   MOCK_METHOD1( Close, void( handle_type ) );

};

struct FunnyGlobal {
  FunnyGlobal() : pimpl() {}
  ~FunnyGlobal() { delete pimpl; }

  // You'd want some protection here rather than just dereferencing.
  // it's the whole point.  I'd suggest using whatever unit test assertion
  // mechanism your framework uses and make it a fatal check.    
  handle_type Open(char const* name) { return pimpl->Open(name); }
  void Close(handle_type h) { pimpl->Close(h); }
private:
   FunnyDoodad * pimpl;

  friend struct FunnyDoodad;

  void register(FunnyDoodad* p) { pimpl = p; }
  void deregister() { pimpl = 0; }
};

FunnyGlobal funnyGlobal;

FunnyDoodad::FunnyDoodad() { funnyGlobal.register(this); }
FunnyDoodad::~FunnyDoodad() { funnyGlobal.deregister(); }

struct TestTraits
{
    typedef HANDLE handle_type;
    static handle_type Open( const char* name ) { return funnyGlobal.Open(name); };
    static void Close( handle_type h ) { funnyGlobal.Close(h); };
};

TEST_CASE(blah)
{
   FunnyDoodad testDoodad;

   ...
}

我想上面可以被模板化，几乎变成了一种模式……也许吧。

于 2012-05-18T18:25:49.027 回答