使用内置数据类型的向前声明 [英] Using forward declarations for built in datatypes

问题描述

我有一个类 Person

code>这样。

  #pragma once 
 
 #include< string& 
 
 class Person 
 {
 public：
 Person（std :: string name，int age）; 
 std :: string GetName（void）const; 
 int GetAge（void）const; 
 private：
 std :: string _name; 
 int _age; 
}; 
  

和一个类 Student p>

  #pragma once 
 
 #include< string> 
 
 class Person; 
 
 class Student 
 {
 public：
 Student（std :: string name，int age，int level = 0）; 
 Student（const Person& person）; 
 std :: string GetName（void）const; 
 int GetAge（void）const; 
 int GetLevel（void）const; 
 private：
 std :: string _name; 
 int _age; 
 int _level; 
}; 
  

在Student.h中，我有一个前进声明 class Person; code>在我的转换构造函数中使用 Person 。精细。但是我在代码中使用 std :: string 时，已经做了 #include< string> 如何使用forward声明这里避免编译错误？是否有可能？

感谢。

解决方案

c $ c> string as

  std :: string _name; 
 // ^^^^^^^^^具体成员
  

整个结构 string 将需要，所以必须声明。您必须 #include< string> 。

---

string 可以省略，例如

  std :: string * _name; 
 // ^^^^^^^^^^指针或参考
  

可以使用向前声明，但我仍然建议您不这样做，因为 std :: string 不是一个简单的结构类型像Person或学生，但涉及许多模板的非常复杂的类型：

  template< class charT，class traits = char_traits< charT& = allocator< charT> > 
 class basic_string {...}; 
 typedef basic_string< char>串; 
  

如果你向前声明错误（例如 class string; ），编译将失败，当你实际使用它，因为冲突的类型。

I understand that wherever possible we shall use forward declarations instead of includes to speed up the compilation.

I have a class Person like this.

#pragma once

#include <string>

class Person
{
public:
    Person(std::string name, int age);
    std::string GetName(void) const;
    int GetAge(void) const;
private:
    std::string _name;
    int _age;
};

and a class Student like this

#pragma once

#include <string>

class Person;

class Student
{
public:
    Student(std::string name, int age, int level = 0);
    Student(const Person& person);
    std::string GetName(void) const;
    int GetAge(void) const;
    int GetLevel(void) const;
private:
    std::string _name;
    int _age;
    int _level;
};

In Student.h, I have a forward declaration class Person; to use Person in my conversion constructor. Fine. But I have done #include <string> to avoid compilation error while using std::string in the code. How to use forward declaration here to avoid the compilation error? Is it possible?

Thanks.

解决方案

Since used string as

std::string _name;
//^^^^^^^^^ concrete member    

the whole structure of string would be needed, so the declaration must be needed. You must #include <string>.

---

Declaration of string can be omitted possible if you write, e.g.

std::string* _name;
//^^^^^^^^^^ pointer or reference

which you could use a forward declaration, but I still recommend you not to do so, because std::string is not a simple structure type like Person or Student, but a very complex type involving many templates:

template<class charT, class traits = char_traits<charT>, class Allocator = allocator<charT> >
class basic_string { ... };
typedef basic_string<char> string;

If you forward declare it wrongly (e.g. class string;), the compilation will fail when you actually use it because of conflicting type.

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