Emi*_*erg 9 c++ templates googletest c++11
假设我有一个容器类Box,它提供内部类const_iterator和iterator.因为我想要一个iterator可以转换为a const_iterator,后者继承自前者:
class Box {
// ...
public:
class const_iterator : public std::iterator<std::random_access_iterator_tag, const int> { /* ... */ };
class iterator : public const_iterator { /* ... */ };
// ...
};
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现在我想使用Google Test测试这些类.让我们断言,begin()并且end()不要返回相同的东西:
const Box a;
EXPECT_NE(a.begin(), a.end());
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跟编译错误打招呼:
no member named 'begin' in 'Box::const_iterator'‘const class Box::const_iterator’ has no member named ‘begin’一些研究让我在Google Test源代码中找到了这个模板(按照扩展文档的链接):
typedef int IsContainer;
template <class C>
IsContainer IsContainerTest(int /* dummy */,
typename C::iterator* /* it */ = NULL,
typename C::const_iterator* /* const_it */ = NULL) {
return 0;
}
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这个模板魔术的结果是,如果EXPECT_*有iterator和const_iterator成员类的参数,那么该类型被假定为容器类.知道了这一点,Google Test可以在预期失败时打印漂亮的人工可读报告,这很不错.
但是,有一些细节:
// Note that we look for both C::iterator and C::const_iterator. The
// reason is that C++ injects the name of a class as a member of the
// class itself (e.g. you can refer to class iterator as either
// 'iterator' or 'iterator::iterator'). If we look for C::iterator
// only, for example, we would mistakenly think that a class named
// iterator is an STL container.
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所以,如果我理解正确,这意味着
Box::const_iterator将自己作为一个名为的成员类const_iterator,并std::iterator作为一个名为的成员类iterator.Box::iterator将自己作为一个名为的成员类iterator和Box::const_iterator一个名为的成员类const_iterator.因此,我的迭代器类看起来都像是Google Test的容器类!
如何设计我的迭代器类以使它们看起来不像容器?
我试过的事情:
std::iterator超类.这解决了隐藏成员类的问题,但它仍然不允许我作为参数传递,除非是.似乎Google Test使用而不是出于某些原因.const_iteratorprivateconst_iteratoriteratora.begin()EXPECT_NEaconstiterator begin()const_iterator begin() conststd::iterator完全删除超类.这是一个坏主意吗?我想我必须std::iterator_traits手动申报,还有什么我不会因为没有延期而失去std::iterator?Box::const_iterator超类.这可能是也可能不是一种选择,因为我必须重新声明我想要重用的方法(例如).Box::iteratorprivateoperator++还有什么我忽略的吗?
#include<iterator>
#include <memory> //unique_ptr<T>
#include <gtest/gtest.h>
class ThreeInts {
std::unique_ptr<int[]> v;
public:
ThreeInts() : v(new int[3]) { v[0] = 0; v[1] = 1; v[2] = 2; };
ThreeInts(int val) : ThreeInts() { v[0] = val; v[1] = val; v[2] = val; };
bool operator==(const ThreeInts& other) const {
return v[0] == other.v[0] && v[1] == other.v[1] && v[2] == other.v[2];
}
class const_iterator : public std::iterator<std::random_access_iterator_tag, const int> {
protected:
int* p;
public:
explicit const_iterator(int* p) : p(p) {}
const_iterator& operator++() { ++p; return *this; }
bool operator==(const const_iterator& rhs) const { return p == rhs.p; }
bool operator!=(const const_iterator& rhs) const { return p != rhs.p; }
int operator*() const { return *p; }
};
class iterator : public const_iterator {
public:
explicit iterator(int* p) : const_iterator(p) {}
int& operator*() const { return *p; }
};
iterator begin() { return iterator(v.get()); }
iterator end() { return iterator(v.get()+3); }
const_iterator begin() const { return const_iterator(v.get()); }
const_iterator end() const { return const_iterator(v.get()+3); }
};
TEST(ThreeInts, ThisTestCompilesAndPrettyFailureMessagesAreShown) {
const ThreeInts a(1), b(2);
ThreeInts c(1), d(2);
EXPECT_EQ(a, b);
EXPECT_EQ(a, c);
EXPECT_EQ(c, d);
}
TEST(ThreeInts, ThisTestCompilesIfTheStdIteratorParentIsPrivate) {
const ThreeInts a;
EXPECT_NE(a.begin(), a.end());
}
TEST(ThreeInts, ThisTestAlsoCompilesIfTheStdIteratorParentIsPrivateButItIsAHassle) {
ThreeInts a;
ThreeInts::const_iterator beg = a.begin();
ThreeInts::const_iterator end = a.end();
//EXPECT_NE(beg, end); // Compile error unless the std::iterator superclass is private
}
TEST(ThreeInts, ThisTestDoesNotCompileEvenIfTheStdIteratorParentIsPrivate) {
ThreeInts a;
//EXPECT_NE(a.begin(), a.end());
}
int main(int argc, char **argv) {
::testing::InitGoogleTest(&argc, argv);
return RUN_ALL_TESTS();
}
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ThreeInts::iterator不应该继承自ThreeInts::const_iterator,而是应该单独实现。
class ThreeInts::iterator : public std::iterator< std::random_access_iterator_tag, int> { ... }
class ThreeInts::const_iterator : public std::iterator< std::random_access_iterator_tag, const int> { ... }
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问题似乎是否则两者都有名为和 的ThreeInts::const_iterator成员(也称为构造函数)。另外,继承不是常量正确的,因为应该只保存一个指针/类似于数据。STL 容器还将两个迭代器分开。const_iteratoriteratoriteratorconst_iteratorconst_iteratorconst
在该代码中,不需要定义迭代器类,只需定义
using iterator = int*;
using const_iterator = const int*;
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