C++ 谷歌测试装置
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Google Test Fixtures
提问by jiake
I'm trying to understand how the Google Test Fixtures work.
我试图了解 Google 测试装置的工作原理。
Say I have the following code:
假设我有以下代码：
class PhraseTest : public ::testing::Test
{
protected:
virtual void SetUp()
{
phraseClass * myPhrase1 = new createPhrase("1234567890");
phraseClass * myPhrase2 = new createPhrase("1234567890");
}
virtual void TearDown()
{
delete *myPhrase1;
delete *myPhrase2;
}
};
TEST_F(PhraseTest, OperatorTest)
{
ASSERT_TRUE(*myPhrase1 == *myPhrase2);
}
When I compile, why does it say "myPhrase1" and "myPhrase2" are undeclared in the TEST_F?
当我编译时，为什么它说 TEST_F 中未声明“myPhrase1”和“myPhrase2”？
回答by Billy ONeal
myPhrase1and myPhrase2are local to the setup method, not the test fixture.
myPhrase1并且myPhrase2是设置方法的本地，而不是测试夹具。
What you wanted was:
你想要的是：
class PhraseTest : public ::testing::Test
{
protected:
phraseClass * myPhrase1;
phraseClass * myPhrase2;
virtual void SetUp()
{
myPhrase1 = new createPhrase("1234567890");
myPhrase2 = new createPhrase("1234567890");
}
virtual void TearDown()
{
delete myPhrase1;
delete myPhrase2;
}
};
TEST_F(PhraseTest, OperatorTest)
{
ASSERT_TRUE(*myPhrase1 == *myPhrase2);
}
回答by Bill
myPhrase1and myPhrase2are declared as local variables in the SetUpfunction. You need to declare them as members of the class:
myPhrase1并且myPhrase2在SetUp函数中被声明为局部变量。您需要将它们声明为类的成员：
class PhraseTest : public ::testing::Test
{
protected:
virtual void SetUp()
{
myPhrase1 = new createPhrase("1234567890");
myPhrase2 = new createPhrase("1234567890");
}
virtual void TearDown()
{
delete myPhrase1;
delete myPhrase2;
}
phraseClass* myPhrase1;
phraseClass* myPhrase2;
};
TEST_F(PhraseTest, OperatorTest)
{
ASSERT_TRUE(*myPhrase1 == *myPhrase2);
}
