变量等末尾的与号 (&)

ampersand (amp;) at the end of variable etc(变量等末尾的与号 (amp;))

问题描述

我是 C++ 菜鸟，我在理解代码中的 C++ 语法方面遇到了问题.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，我理解它作为引用、地址和逻辑运算符的一般用法...

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

& 是什么意思?参数末尾的运算符?

What is the meaning of an & operator at end of parameter?

friend int date_ok( const date& );

谢谢

感谢您的回答.如果我理解正确的话，变量名被简单地省略了，因为它只是一个原型.对于原型，我不需要变量名，它是可选的.对吗?

Thanks for the answers. If I have understood this correctly, the variable name was simply omitted because it is just a prototype. 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，我肯定需要变量名，对吗?

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

推荐答案

const date& 被方法 date_ok 接受意味着 date_ok接受 const date 类型的引用.它的工作原理与指针类似，只是语法稍微有点……含糖

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

在您的示例中，int* Y = &x 使 Y 成为 int * 类型的指针，然后为其分配地址x.当我想更改Y 指向的地址中的任何内容"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

相反，我现在使用参考

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

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