在变量结束处的&符号（&）等 [英] ampersand (&) at the end of variable etc

问题描述

我是一个C ++ noob，我有一个问题，在代码中理解c ++语法。现在我很困惑。

I am a C++ noob and i've a problem of understanding c++ syntax in a code. Now I am quite confused.

class date
{
private:
int day, month, year;
int correct_date( void );
public:
void set_date( int d, int m, int y );
void actual( void );
void print( void );
void inc( void );
friend int date_ok( const date& );
};

关于'&'字符，我理解它的一般用法作为参考，地址和逻辑运算符...

Regarding to the '&' character, I understand its general usage as a reference, address and logical operator...

例如int * Y =& X

for example int *Y = &X

;

friend int date_ok( const date& );

感谢

edit:

感谢回答者。
如果我正确理解了这一点，变量名只是简单地省略了，因为它只是原型。对于原型我不需要变量名，它是可选的。这是正确的吗？

Thanks for the answesers. If I have understood this correctly, the variable name simply was omitted there because it is just the prototype. And for the prototype i don't need the variable name, it's optional. Is that correct?

但是，对于函数的定义，我需要一定的变量名，对吗？

However, for the definition of the function I need definitely the variable name, right?

推荐答案

const日期& 被方法 date_ok 接受意味着 date_ok 引用 const date 。它的工作类似于指针，除了语法稍微更多.. sugary

const date& being accepted by the method date_ok means that date_ok takes a reference of type const date. It works similar to pointers, except that the syntax is slightly more .. sugary

在你的例子中， int * Y =& code>使 int 类型的指针 Y 。然后为其分配 x 的地址。当我想改变 Y 所指向的地址的值时，我说 * Y = 200;

in your example, int* Y = &x makes Y a pointer of type int. and then assigns it the address of x. And when I would like to change the value of "whatever it is at the address pointed by Y" I say *Y = 200;

所以，

int x = 300;
int *Y = &x;
*Y = 200; // now x = 200
cout << x; // prints 200

而现在我使用参考

int x = 300;
int& Y = x;
Y = 200; // now x = 200
cout << x; // prints 200

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