Question

我有以下法典:

struct date {
    int day;
    int month;
    int year;
};

class mydateclass {
    public: 
        int day;
        int month;
        int year;
};

mydateclass date;

date.day;

正在提及哪些<条码>变量? 。

Answer 1

标题声明称为“日期”。在<代码>mydate 日期;之前没有物体date>。因此,“召唤”并非野心。

如果你想以这种方式制造一个物体,那就是:

struct datestruct { int day; int month; int year; } date;

如果你这样做,你的汇编者应在<条码>日报上报;,作为已经存在的标语。

请注意,如果你想与某类/构件的某一成员(如固定成员)打交道,那么你就不需要。例如:

struct date { static int day; }; date::day;

Answer 2

There is no ambiguity.
The code defines a structure, but it doesn t create an instance of it. When you write date.day, you are referring to a variable, and there is just a variable named date.

Answer 3

由于<代码><>在本代码中被宣布为<代码>mydatenal<>/code>的例,该例标有您可以通过<<代码>><>>>查询的成员类型。

稍加修改的法典的快速测试表明,变数以相同名称掩盖了结构:

struct date {
    int day;
    int month;
    int year;
};

class mydateclass {
    public:
        int day;
        int month;
        int year;
};

int main()
{
    mydateclass date;

    date.day;

    date other_date;  //error here, because "date" does not name a type in this context
    ::date yet_another_date; //okey, because "::" means the name in question is in global scope
                             //whereas the variable named "date" is local to main
}

void foo()
{
    date other_date;  //ok here, because there is no variable called "date" in scope here
}

无论你有结构或等级,都不会对问题有任何区别。

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