非常量静态成员的错误使用方法 静态成员只跟类有关,是所有对象共有的属性,如果在类中初始化则意味着每个对象都初始化一次,这样就需要非静态成员了。非常量静态成...
C.134: Ensure all non-const data members have the same access level C.134:确保所有非常量数据成员具有相同的访问权限 Reason...If the non-const data members don't have the same access level, the type is confused about what it's...To make them non-private and non-const would mean that the object can't control its own state: An unbounded...An encapsulated object may contain something like non-const debug instrumentation that isn't part of...Enforcement(实施建议) Flag any class that has non-const data members with different access levels.
(1)不能将const 类型的对象传递给non-const 类型的引用。...此时通过const 引用不能修改对象,但对象可以通过non-const 引用被修改。...即,const 类型的对象不能直接被nonconst 类型的指针所指示(同2);可以使用const 类型的指针指向non-const 对象(同3)。...5.可以将相同类型(包括常量性)的const 指针值赋给non-const 指针。...但自定义类型 则为non-const 性质。
public: // BAD, duplicates logic Bar& get_bar() { /* complex logic around getting a non-const...Normally, you can just have the non-const function call the const function....class Foo { public: // not great, non-const calls const version but resorts to const_cast Bar...my_bar; }; Although this pattern is safe when applied correctly, because the caller must have had a non-const...} private: Bar my_bar; template // good, deduces whether T is const or non-const
F.17: For "in-out" parameters, pass by reference to non-const(输入/输出参数传递非常量引用) 译者注:in-out指的是向函数传递输入信息的同时又从函数获取输出信息的参数...Enforcement(实施建议) (Moderate) ((Foundation)) Warn about functions regarding reference to non-const parameters...(Simple) ((Foundation)) Warn when a non-const parameter being passed by reference is moved.
execution access the same instance of shared_ptr without synchronization and any of those accesses uses a non-const...首先来看一下 std::shared_ptr 的所有成员函数,只有前 3 个是 non-const 的,剩余的全是 const 的: 成员函数 是否 const operator= non-const...reset non-const swap non-const get const operator*、operator-> const operator const use_count const unique...例 2 是有数据竞争存在的,因为所有 thread 都共享了同一个 test 的引用,根据刚才的结论 2,对于同一个 std::shared_ptr 对象,多线程访问 non-const 的函数是非线程安全的
版本,该函数可能用来读取,也可能用来写一个字符,C++编译器无法告诉我们operator[]被调用时是用于写还是取,所以我们必须假设所有的non-const operator[]的调用都用于写。...//重载[]运算符,针对non-const Strings char& String::operator[](size_t index){ if (value->refCount>1){...这个不问题不限于指针,如果有人以引用的方式将String的non-const operator[]返回值存储起来,也会发生同样的问题。 解决这种问题主要有三种方法。 (1)忽略之。...一开始,我们先树立此标志为true,表示对象可被共享,但只要non-const operator[]作用于对象值时就将标志清楚。一旦标志被设为false,那么数据实体可能永远不会再被共享了。...对于Non-const operator[]是唯一将shareable设为false者,其实现代码可为: char& String::operator[](size_t index){ if
char* p); // g does not modify *p Note(注意) It is not inherently bad to pass a pointer or reference to non-const...属性的转换 Enforcement(注意) Flag function that does not modify an object passed by pointer or reference to non-const
because getx was not marked const } Note(注意) It is not inherently bad to pass a pointer or reference to non-const...const member function to change the value of mutable members and the value of objects accessed through non-const...考虑使用指向实现的指针技术获得稳定的ABI Enforcement(实施建议) Flag a member function that is not marked const, but that does not perform a non-const
., unique_ptr), or passing it in a reference to non-const target object to fill (to be used as an out-parameter...Enforcement(实施建议) Flag reference to non-const parameters that are not read before being written to and...To fix: Remove const to return a non-const value instead. 警告返回常数值的状况。修改方法:去掉常量修饰,返回一个非常量。
template const T* RCIPtr::operator->() const { return counter->pointee; } //重载*运算符,non-const...counter->pointee = new T(*oldValue); counter->addReference(); } } //重载->运算符,non-const...typename T> T* RCIPtr::operator->(){ makeCopy(); return counter->pointee; } //重载*运算符,non-const...第二是RCIPtr将operator->和operator*重载了,如此一来只要有non-const access发生在被指物身上,copy-on-write(写时复制)就会被执行。
immutable Con.1:默认情况下使对象不可修改 Reason(原因) Immutable objects are easier to reason about, so make objects non-const...Enforcement(实施建议) Flag non-const variables that are not modified (except for parameters to avoid many
pointer, non-const data const char* p = myName; // non-const pointer, const data char* const p...对象 std::cout << ms[0]; // 调用 non-const MyString::operator[] ms[0] = 'x'; // 没问题,写一个 non-const...虽然语法有一点点奇特,但「运用 const 成员函数实现 non-const 孪生兄弟 」的技术是值得了解的。...需要注意的是:我们可以在 non-const 成员函数调用 const 成员函数,但是不可以反过来,在 const 成员函数调用 non-const 成员函数调用,原因是对象有可能因此改动,这会违背了...当 const 和 non-const 成员函数有着实质等价的实现时,令 non-const 版本调用 const 版本可避免代码重复。
+ lv2; // legal, rvalue ref extend lifecycle rv2 += "string"; // legal, non-const...return 0; } 一个很有趣的历史遗留问题 #include int main() { // int &a = std::move(1); // illegal, non-const
const 变量的操作: const int a = 10; a = 11; // 错误:不能为 const 变量赋新值 int &b = a; // 错误:不能为 cosnt 变量绑定 non-const...引用 const int &c = a; // 正确 int *d = &a; // 错误:不能为 const 变量绑定指向 non-const 的指针 const int *e = &a; //
int function() {return mem;} private: int mem; }; const object(datamember 不可变动) non-const...object(datamem 可变动) const member functions(保证 datamenber 不变) YES YES non-const member functions(不保证...datamember 不变) 无法调用 YES 当成员函数的 const 和 non-const 版本同时存在时,const Object 只会调用 const 版本,non-const Object...只会调用 non-const 版本 ---- 关于 new , delete. array new , array delete.
Enforcement(实施建议) Flag unmodified non-const variables. 标记不会被修改的非常量。
看代码 struct Functor { void operator()() { std::cout << "Non-const\n"; } void operator()() const...Const all the way f(); // Prints "Non-const" 问题就在于function的表现,复制Functor...也就是这样 std::function f = ftor; f(); // prints "Non-const" const std::function f
size_type pos ) const; 原因如下: void f(std::vector & v1, std::vector const & v2) { //v1 is non-const...vector //v2 is const vector auto & x1 = v1[0]; //invokes the non-const version auto & x2...= v2[0]; //invokes the const version v1[0] = 10; //okay : non-const version can modify the object
mutable 用于修饰 (1)non-static (2)class members of (3)non-reference (4)non-const (5)type。
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