Google Mock - 使用:: testing :: An()

时间:2011-02-20 10:51:33

标签: testing googlemock

使用以下Google Mock声明时收到编译错误:

EXPECT_CALL(some_object, someFunction(1,An<AStructIDefined>()))
    .Times(2);

错误是:

1>ClCompile:
1>  TestMyClass.cpp
1>TestMyClass.cpp(189): error C2664: 'mynamespace::MockMyClassClient::gmock_someFunction' : cannot convert parameter 2 from 'testing::Matcher<T>' to 'const testing::Matcher<T> &'
1>          with
1>          [
1>              T=mynamespace::AStructIDefined
1>          ]
1>          and
1>          [
1>              T=const mynamespace::AStructIDefined &
1>          ]
1>          Reason: cannot convert from 'testing::Matcher<T>' to 'const testing::Matcher<T>'
1>          with
1>          [
1>              T=mynamespace::AStructIDefined
1>          ]
1>          and
1>          [
1>              T=const mynamespace::AStructIDefined &
1>          ]
1>          No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called

我做错了什么?


更新

我正在使用VS2010。

someFunction的声明是:

virtual void someFunction( long long ll, const AStructIDefined& a_struct);

An()是Google Mock wildcard matcher,其定义如下:

// Creates a matcher that matches any value of the given type T.
template <typename T>
inline Matcher<T> An() { return A<T>(); }

结构的简化但有代表性的版本是:

namespace mynamespace {

class ABaseCLass
{
public:
    virtual ~ABaseCLass(){};
    virtual bool isValid() const = 0;
};

struct AStructIDefined : public ABaseCLass
{
public:
    OrderStatusReport(SomeEnum1 e_, int i_, double d_);

    SomeEnum1 e;
    int i;
    double d;

    const std::string toString() const;
    bool isSane() const;
    bool operator== (const SomeEnum1& ref_) const;
    double getD() const;
    int getI() const;
    bool isCondition() const;
};

} // namespace mynamespace

1 个答案:

答案 0 :(得分:5)

解决方案是改变声明:

EXPECT_CALL(some_object, someFunction(1,An<AStructIDefined>()))
    .Times(2);

EXPECT_CALL(some_object, someFunction(1,An<const AStructIDefined &>()))
    .Times(2);

C ++隐式在函数参数上强制转换const和引用&,但google mock的声明似乎要求函数的签名中出现类型,而不是作为函数参数提交的类型。