我在代码的int main()
部分调用类构造函数时遇到了一些困难。
它是一个构造函数,以字符串数组作为参数。
我知道,在调用构造函数时,可以设置默认参数或根本不设置参数,并且需要一个对象来调用构造函数(以及我们想要在其中提供的参数)。但我仍然不明白如何称呼它,尽管我尝试了很多不同的方法。
这是我的代码:
#include <iostream>
#include <string>
using namespace std;
enum player_position{ GoalKeeper, Midfielder, Defender, Striker };
class Football_Player
{
private:
string Name;
int Age;
int Points;
player_position Ppos;
public:
Football_Player(string _name = "aname", int _age = 20, int _points = 50, player_position _ppos = Striker)
{
Name = _name;
Age = _age;
Points = _points;
Ppos = _ppos;
}
Football_Player(string str[4]) // <---- "This Constructor is the one , i can't seem to call into the main()."
{
cin >> str[0];
Name = str[0];
cout << Name;
cout << str[0];
int a = atoi(str[1].c_str());
cout << a;
int b = atoi(str[2].c_str());
cout << b;
str[3] = Ppos;
}
};
int main()
{
// Please don't take any of the info as biased, these are just random.
Football_Player("Messi", 20, 50, Striker);// This one is the previous constructor with the default arguments and this one seems to be working.
Football_Player (); // Trying to call that constructor
Football_Player object1("Messi"); // Trying to call that constructor
Football_Player object2("Ronaldo", 25, 50, Striker); // Again trying to call that Constructor
Football_Player object3(str[0]); // And Again . . . .
system("pause");
return 0;
}
答案 0 :(得分:1)
当您在第一个CTor中声明了4个默认值时,您的通话Football_Player object1("Messi");
实际上会调用该默认值,并离开
age = 20
points = 50
position = Striker
你要么“必须给所有参数,要么没有”,这是完全错误的。对于你给出的所有论点,位置很重要。在您的示例中:如果您提供2个参数,则提供名称和年龄。没有办法只给出分数和位置。此外,无法调用Football_Player object1("Messi",,,Midfield);
这样的电话。
第二个构造函数总是需要一个包含4个字符串的数组。没有更多,没有更少。但我建议删除那个,因为没有技巧你也可以给它一个指向字符串的指针,导致崩溃。