我只是在上课,所以我正在尝试一些基本的东西。我有一个名为Month的类,如下所示。对于我的第一次测试,我想提供1到12之间的数字并输出月份名称即。 1 = 1月。
class Month
{
public:
Month (char firstLetter, char secondLetter, char thirdLetter); // constructor
Month (int monthNum);
Month();
void outputMonthNumber();
void outputMonthLetters();
//~Month(); // destructor
private:
int month;
};
Month::Month()
{
//month = 1; //initialize to jan
}
void Month::outputMonthNumber()
{
if (month >= 1 && month <= 12)
cout << "Month: " << month << endl;
else
cout << "Not a real month!" << endl;
}
void Month::outputMonthLetters()
{
switch (month)
{
case 1:
cout << "Jan" << endl;
break;
case 2:
cout << "Feb" << endl;
break;
case 3:
cout << "Mar" << endl;
break;
case 4:
cout << "Apr" << endl;
break;
case 5:
cout << "May" << endl;
break;
case 6:
cout << "Jun" << endl;
break;
case 7:
cout << "Jul" << endl;
break;
case 8:
cout << "Aug" << endl;
break;
case 9:
cout << "Sep" << endl;
break;
case 10:
cout << "Oct" << endl;
break;
case 11:
cout << "Nov" << endl;
break;
case 12:
cout << "Dec" << endl;
break;
default:
cout << "The number is not a month!" << endl;
}
}
这是我有问题的地方。我想将“num”传递给outputMonthLetters函数。我该怎么做呢?该函数是无效的,但必须有一些方法将变量放入类中。我必须公开“月份”变量吗?
int main(void)
{
Month myMonth;
int num;
cout << "give me a number between 1 and 12 and I'll tell you the month name: ";
cin >> num;
myMonth.outputMonthLetters();
}
答案 0 :(得分:4)
你可能想要做的是这样的事情:
int num;
cout << "give me a number between 1 and 12 and I'll tell you the month name: ";
cin >> num;
Month myMonth(num);
myMonth.outputMonthLetters();
请注意,myMonth在需要之前不会被声明,并且在您确定要查找的月份数后,将调用采用月份编号的构造函数。
答案 1 :(得分:1)
尝试使用方法
上的参数void Month::outputMonthLetters(int num);
比你能做的更多:
Month myMonth;
int num;
cout << "give me a number between 1 and 12 and I'll tell you the month name: ";
cin >> num;
myMonth.outputMonthLetters(num);
我不是C ++大师,但是你不必创建Month的实例吗?
答案 2 :(得分:0)
更改您的
void Month::outputMonthLetters()
到
static void Month::outputMonthLetters(int num)
{
switch(num) {
...
}
}
即。向方法添加参数,并(可选)使其成为静态。但这不是一个很好的例子,从一开始......