Question

我正在尝试根据这些说明创建一个Temperature类：

考虑一个叫做温度的类。该类由一个名为degrees的double和一个名为type的字符串组成。该类有一个默认的构造函数。该类还有一个构造函数，它接受只接受degrees变量参数的参数。它有一个名为get_temperature（）的访问器，它返回一个double。它有一个过载的bool运算符＆lt;它将另一个温度对象T作为参数。它还有一个名为set_type的mutator成员函数，它将一个字符串作为参数并且不返回任何内容。写出类温度的声明。适当时使用const关键字。

#include <iostream>
#include <string>
#include <time.h>

using namespace std;

class Temperature {
public:
    Temperature (double degrees_x){
        double degrees = degrees_x;
    }
    void set_type(string type_x){
        string type = type_x;}
        double get_temperature() const;
    bool operator < (Temperature T) const;
private:
    double degrees;
    string type;


};

Answer 1

可以/应该纠正的一些事情。

运营商＆lt;参数可以是const引用

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避免＆＃34;使用命名空间＆＃34;在头文件中 - 污染全局命名空间

行：

bool operator < (const Temperature& T) const;

定义一个局部变量，您可以根据传递给构造函数和set_type方法的参数来赋值。因此，对象成员字段未按预期初始化。对于构造函数，您可以简单地添加

double degrees = degrees_x;
string type = type_x;

至于set_type方法：

Temperature (double degrees_x) : degrees(degrees_x), type(""){}

此方法也可以采用const＆amp;参数而不是字符串值

此外，这里似乎没有两个包含（time.h和iostream）。

Answer 2

这是一个修改后的版本，其中包含一些有关更改的内联注释。

#include <iostream>
#include <string>
#include <time.h>

using namespace std;

class Temperature {
public:
    Temperature (double degrees_x) : 
      degrees(degrees_x)      // this sets the value of the object variable
                              // it means that we can use const double degrees later
    {}

    // Note here that the signature has changed to be a const & 
    void set_type(const string& type_x) {
        type = type_x;         //sets the object variable type to the value.
        // one could alternatively use the non-const, and then std::move.

        //string type = type_x;  but this creates a local variable, and assigns the value... not what you want.
    }

    // good uses of const here for function declarations.
    double get_temperature() const;
    bool operator < (const Temperature& t) const; // but this should be const & argument, since you shouldn't change it
    // as a style note, T is a common name for template parameters, so I'd suggest naming it t rather than T
private:
    const double degrees;  // this never changes after the constructor
    string type;           // this however can, so mustn't be const
};

Answer 3

    #include <iostream>
#include <string>
#include <time.h>

using namespace std;

class Temperature {
    //no need to declare private members as private, the default is private
    double degrees;
    string type;
public:
    Temperature (double degrees_x){
        degrees = degrees_x;
    }
    //when setting the type, using 'set_type(string type_x) is inefficient, as type_x will be copied if given an l-value, and then copied again into 'type'
    //getting 'type_x' as const& will make sure you make a single copy, which is in 'string type = type_x' when the copy assigment of the class 'string' is called
    void set_type(const string& type_x){
        string type = type_x;
    }
    double get_temperature() const;
    //the most reasonable thing to do, as already mentioned, is passing T as const&, since you probably won't modify the second operand when checking if T1<T2
    bool operator < (const Temperature &T) const;
};

类。有人能告诉我，我是否正确地进行了这项运动？

3 个答案: