Question

我正在玩gmock，我有一个人为的例子，我正在用它来学习它的细微差别。我有一个问题，调用我所期望的隐式复制构造函数：

// mock_word.h
class MockWord : Word {
public:
    MockWord(const std::string word) : Word(word) {};
    MOCK_METHOD0(pigLatinify, std::string(void));
};

// strings.h
template <typename Word>
class Strings {
...
private:
    std::vector<Word>* words = new std::vector<Word>();
public:
    // This should call the implicit copy constructor
    void addWord(const Word word) {
        this->words->push_back(word); 
    };
...
};

// strings_test.cpp
class StringsTest : public ::testing::Test {
protected:
    Strings<MockWord>* strings;
public:
    virtual void SetUp() {
        strings = new Strings<MockWord>();
    };
    virtual void TearDown() {
        delete strings;
    };
};

TEST_F(StringsTest, StringIsAllPigLatinifiedNicely) {
    MockWord mockWordA("beast");
    MockWord mockWordB("dough");

    // Set some expectations for the Mock
    EXPECT_CALL(mockWordA, pigLatinify()).Times(AtLeast(1)); 
    EXPECT_CALL(mockWordB, pigLatinify()).Times(AtLeast(1)); 

    strings->addWord(mockWordA);
    strings->addWord(mockWordB);
    ...
};

现在，如果我将mockWordA和mockWordB从自动变量转换为指针，我可能有这个complile并且工作，但那不是我想要提供的接口。

我得到的确切错误是：

error: call to implicitly-deleted copy constructor of 'MockWord'
    strings->addWord(mockWordA);
                     ^~~~~~~~~
mock_word.h:11:9: note: copy constructor of 'MockWord' is implicitly     deleted because field 'gmock0_pigLatinify_11' has a deleted copy     constructor
    MOCK_METHOD0(pigLatinify, std::string(void));

Answer 1

我通过将我的函数接口更改为Strings::addWord(Word* word)解决了这个问题，这对我来说似乎是限制性的，因为我不想使用带有这样一个基本示例的指针。

gmock调用隐式删除的复制构造函数

1 个答案: