[英]access a member of class using vector
I have a class defined as 我有一个定义为的类
typedef std::string Name;
typedef int Age;
typedef std::string Level;
class Employee
{
public:
// Employee (arguments);
// virtual ~Employee ();
void name(Name const & name) { name_ = name; }
Name name() const { return name_; }
void age(Age const & age) { age_ = age; }
Age age() const { return age_; }
void level(Level const & level) { level_ = level; }
Level level() const { return level_; }
private:
Name name_;
Age age_;
Level level_;
};
std::vector<Employee> read_file(std::string filename);
std::vector<Employee> employees = read_file("data.txt");
std::cout << employees.size() << std:: endl;
for(std::vector<Employee>::iterator it = employees.begin(); it != employees.end(); ++it)
{
std::cout << *it << std::endl;
}
Is there a way for me to access the members of the class Employee using this vector defined above? 有没有办法让我使用上面定义的这个向量访问Employee类的成员? I want to construct a
map
container with level of the employee as the key value. 我想构建一个以雇员级别作为键值的
map
容器。
如果要从迭代器访问Employee
成员,可以使用成员访问运算符->
:
the_map[it->level()] = blah;
const Name name = it->name();
const Age age = it->age();
const Level level = it->level();
or 要么
const Name name = employees[i].name();
const Age age = employees[i].age();
const Level level = employees[i].level();
will work fine. 会很好的。
I would, however, strongly recommend you return each of the above items as references as it will generally be a lot faster than making a copy of the object. 但是,我强烈建议您将上述每个项目作为参考返回,因为它通常比制作对象的副本快得多。
ie 即
class Employee
{
public:
// Employee (arguments);
// virtual ~Employee ();
void name(Name const & name) { name_ = name; }
Name& name() { return name_; }
const Name& name() const { return name_; }
void age(Age const & age) { age_ = age; }
Age& age() { return age_; }
const Age& age() const { return age_; }
void level(Level const & level) { level_ = level; }
Level& level() { return level_; }
const Level& level() const { return level_; }
private:
Name name_;
Age age_;
Level level_;
};
Then you can access the values by reference as follows: 然后,您可以通过引用访问值,如下所示:
const Name& name = it->name();
const Age& age = it->age();
const Level& level = it->level();
It also means you can change the values like this: 这也意味着你可以改变这样的值:
it->name() = "Goz";
it->age() = 33;
it->level() = "Programmer";
Unless I'm misinterpreting your question, this is simple enough. 除非我误解你的问题,否则这很简单。
it
is an iterator over the vector of Employee
. it
是Employee
向量的迭代器。 *it
is the Employee
*it
是Employee
it->
lets you access members of that Employee
it->
允许您访问该Employee
成员
it->name()
it->name()
So, if you want to build a map of employee level to Employee
, you can do this: 因此,如果您要为
Employee
构建员工级别的地图,您可以这样做:
std::map<Level, Employee> employeeMap;
for(std::vector<Employee>::iterator it = employees.begin();
it != employees.end();
++it)
{
employeeMap[ it->level() ] = *it;
}
Now you have your employeeMap, which maps the employee's Level
to the Employee
现在您拥有了employeeMap,它将员工的
Level
映射到Employee
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.