我想为 AutoData 类编写一个打印函数,其中包含有关汽车的信息。通过这个打印函数,我理想地希望打印出包含许多不同类对象的向量。我已经为对象的每个元素编写了 get 函数,但我仍然有点不确定如何使用这些函数来编写函数以以下格式打印数据:
mpg:cylinders:displacement:horsepower:weight:acceleration:modelYear:origin:carName
例如:
10.0:8:360.0:215.0:4615.:14.0:70:1:ford f250
10.0:8:307.0:200.0:4376.:15.0:70:1:chevy c20
11.0:8:318.0:210.0:4382.:13.5:70:1:dodge d200
班级是:
#include <string>
#include <vector>
#include <iostream>
using namespace std;
class AutoData {
public:
AutoData()
{
mpg = 0;
cylinders = 0;
displacement = 0;
horsepower = 0;
weight = 0;
acceleration = 0;
modelYear = 0;
origin = 0;
carName = "";
}
AutoData( const AutoData & rhs)
{
setAuto(rhs.mpg, rhs.cylinders, rhs.displacement, rhs.horsepower, rhs.weight, rhs.acceleration, rhs.modelYear, rhs.origin, rhs.carName);
}
void setAuto(float mp, int cy, float di, float ho, float we, float ac, int mo, int o, string ca)
{
mpg = mp;
cylinders = cy;
displacement = di;
horsepower = ho;
weight = we;
acceleration = ac;
modelYear = mo;
origin = o;
carName = ca;
}
const float & getmpg( ) const
{
return mpg;
}
const int & getcylinders( ) const
{
return cylinders;
}
const float & getdisplacement( ) const
{
return displacement;
}
const float & gethorsepower( ) const
{
return horsepower;
}
const float & getweight( ) const
{
return weight;
}
const float & getacceleration( ) const
{
return acceleration;
}
const int & getmodelYear( ) const
{
return modelYear;
}
const int & getorigin( ) const
{
return origin;
}
const string & getcarName( ) const
{
return carName;
}
bool operator == (const AutoData & rhs ) const
{
if( getmpg( ) == rhs.getmpg( ) )
{
return gethorsepower( ) == rhs.gethorsepower( );
}
else
{
return false;
}
}
bool operator > ( const AutoData & rhs ) const
{
if( rhs.getmpg( ) > getmpg( ) )
{
return true;
}
else if( getmpg( ) == rhs.getmpg( ) )
{
if( rhs.gethorsepower( ) > gethorsepower( ) )
{
return true;
}
}
else
{
return false;
}
}
private:
float mpg;
int cylinders;
float displacement;
float horsepower;
float weight;
float acceleration;
int modelYear;
int origin;
string carName;
};
任何人都可以提供的任何帮助/建议将非常感激!谢谢