c++ - C ++在类中正确使用ostream并传递参数?
问题描述
我最近开始学习 c++,在我的一生中,我似乎无法获得ostream
在 a中使用的语法class
以及我应该传递哪些参数。这是代码:
这是有问题的课程:
#include <iostream>
#include <string>
using namespace std;
class Pokemon{
friend ostream& operator<<(ostream&, Pokemon);
public:
string name, level, cp;
Pokemon(string x="Pikachu", string y="5", string z="1000"){
name = x;
level = y;
cp = z;
}
Pokemon name(){
return this->name;
}
Pokemon level(){
return this->level;
}
Pokemon cp(){
return this->cp;
}
Pokemon display_stats(){
cout << this-> name << "stats are:" << endl;
cout << " " << "Attack: 2716.05" << endl;
cout << " " << "Defence: 1629.63" << endl;
cout << " " << "HP: 1086.42" << endl;
}
};
template<typename TYPE> //i dont understand this and the things i've written down here are only based on samples i've seen
ostream& operator<<(ostream& os, Pokemon & c){
os << "The level of " << c.name << " is" << c.level << " with cp of " << c.cp;
}
正如你所看到的,我已经尝试过构建这个ostream
东西,但我并不真正理解它是如何工作的。这是我的主要功能:
int main()
{
Pokemon a, b, c, d;
a = Pokemon();
b = Pokemon("Weezing");
c = Pokemon("Nidoking", 100);
d = Pokemon("Mewtwo", 50, 5432.1);
cout << a << endl;
cout << b << endl;
cout << c << endl;
cout << d << endl;
cout << "Jessie: You are no match to me! Go " << b.name << "!" << endl;
cout << "Gary: Go lvl " << c.level << " " << c.name << "! Crush them" << endl;
cout << "Ash: " << a.name << " can do it even thouh he is only level " << a.level << endl;
cout << "Jessie: Hahaha! My " << b.name << " CP is " << b.cp << endl;
cout << "Gary: "<< c.name << " CP is " << c.cp << endl;
cout << "Ash: " << a.name << " CP is " << a.cp << endl;
cout << "Giovanni: Behold " << d.name << " is here." << endl;
d.display_stats();
return 0;
}
我收到以下错误:
no instance of constructor "Pokemon::Pokemon" matches the argument list -- argument types are: (const char [9], int) //on line c = Pokemon("Nidoking", 100);
no instance of constructor "Pokemon::Pokemon" matches the argument list -- argument types are: (const char [7], int, double) //on line d = Pokemon("Mewtwo", 50, 5432.1);
解决方案
您所有的Pokemon
类方法都返回错误的类型。而且您main()
根本没有正确调用任何方法。
改变你的Pokemon
班级看起来更像这样:
#include <iostream>
#include <string>
using namespace std;
class Pokemon {
private:
string m_name;
int m_level;
double m_cp;
friend ostream& operator<<(ostream&, const Pokemon&);
public:
Pokemon(string x="Pikachu", int y=5, double z=1000) {
m_name = x;
m_level = y;
m_cp = z;
}
string name() const {
return m_name;
}
int level() const {
return m_level;
}
double cp() const {
return m_cp;
}
void display_stats() const {
cout << m_name << " stats are:" << endl;
cout << " " << "Attack: 2716.05" << endl;
cout << " " << "Defense: 1629.63" << endl;
cout << " " << "HP: 1086.42" << endl;
}
};
ostream& operator<<(ostream& os, const Pokemon &c) {
os << "The level of " << c.m_name << " is " << c.m_level << " with cp of " << c.m_cp;
return os;
}
然后更改main()
为看起来更像这样:
int main()
{
Pokemon a;
Pokemon b("Weezing");
Pokemon c("Nidoking", 100);
Pokemon d("Mewtwo", 50, 5432.1);
cout << a << endl;
cout << b << endl;
cout << c << endl;
cout << d << endl;
cout << "Jessie: You are no match to me! Go " << b.name() << "!" << endl;
cout << "Gary: Go lvl " << c.level() << " " << c.name() << "! Crush them" << endl;
cout << "Ash: " << a.name() << " can do it even though he is only level " << a.level() << endl;
cout << "Jessie: Hahaha! My " << b.name() << " CP is " << b.cp() << endl;
cout << "Gary: " << c.name() << " CP is " << c.cp() << endl;
cout << "Ash: " << a.name() << " CP is " << a.cp() << endl;
cout << "Giovanni: Behold " << d.name() << " is here." << endl;
d.display_stats();
return 0;
}