blocks|key|3311479|text|我假设你有一个打字错误，并且这行实际上是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3311480|X+&operator+=(int+z)+{|code-block|syntax|javascript|3311481|&意味着返回类型是一个引用；您应该将其读作返回X+&类型的函数operator+=。|offset|length|style|CODE|3311482|entityMap^0|0|0|0|1|N|3|V|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

I'm going to assume you have a typo and that the line actually reads:

<pre><code>X &amp;operator =(int z) {
</code></pre>

The <code>&amp;</code> means that the return type is a reference; you should read this as function <code>operator =</code> that returns a type of <code>X &amp;</code>.

blocks|key|3311494|text|我认为这是X&+operator=+(int+z)，也就是说=操作符的重载，它接受一个整数，并返回一个对X的引用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3311495|entityMap^0|5|K|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

It's <code>X&amp; operator= (int z)</code> I think, that is to say a overload of the = operator which takes an int and return a reference to X.

blocks|key|3311523|text|也许，您的代码应该是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3311524|class+X+{
public:
++++int+temp;
private+:
++++//Some+functions+here+
++++X+&operator+=(int+z)+
++++{
++++++++temp+=+z;+
++++++++return+*this+;
++++}
};|code-block|syntax|javascript|3311525|然后你和operator=打交道，而不是和&operator打交道|offset|length|style|CODE|3311526|您的operator+=返回它所应用的对象的引用。|3311527|entityMap^0|0|0|4|9|L|9|0|2|A|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

Probably, your code should look like that:

<pre><code>class X {
public:
 int temp;
private :
 //Some functions here 
 X &amp;operator =(int z) 
 {
 temp = z; 
 return *this ;
 }
};
</code></pre>

And than you deal with <code>operator=</code>, not <code>&amp;operator</code>

Your <code>operator =</code> returns reference to the object it applied to.

blocks|key|3311577|text|如果你稍微改变一下间距，意思可能会更清楚：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3311578|X&+operator=+(int+z)|code-block|syntax|javascript|3311579|这是赋值操作符operator=的重载，它接受一个int参数，并返回一个对class+X的引用。|offset|length|style|CODE|3311580|您可以使用它为对象分配一个整数值：|3311581|X+x;
x+=+42;+//+calls+the+overloaded+operator|3311582|返回值允许您链接赋值：|3311583|X+x1,x2;
x1+=+x2+=+42;+++//+equivalent+to+`x2+=+42;+x1+=+x2;`
(x1+=+x2)+=+42;+//+equivalent+to+`x1+=+x2;+x1+=+42;`|3311584|entityMap^0|0|0|7|9|P|3|11|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|V|$]]

The meaning might be clearer if you change the spacing slightly:

<pre><code>X&amp; operator= (int z)
</code></pre>

That's an overload of the assignment operator, <code>operator=</code>, which takes an <code>int</code> argument, and returns a reference to <code>class X</code>.

You can use it to assign an integer value to an object:

<pre><code>X x;
x = 42; // calls the overloaded operator
</code></pre>

The return value allows you to chain assignments:

<pre><code>X x1,x2;
x1 = x2 = 42; // equivalent to `x2 = 42; x1 = x2;`
(x1 = x2) = 42; // equivalent to `x1 = x2; x1 = 42;`
</code></pre>

blocks|key|5423634|text|这是赋值操作符的定义:它可能看起来像这样更好地格式化(行外定义)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|5423635|X+&+X::operator=(int+z)
{
++++temp+=+z;
++++return+*this;
}|code-block|syntax|javascript|5423636|它允许您编写类似如下的代码：|5423637|X+a,+b,+c(10);
a+=+b+=+c;|5423638|每个分配子表达式的值都是对受让人的引用。|5423639|运算符需要在类定义中声明：|5423640|class+X
{
public:
++++X+&+operator=(int);
++++//+...
};|5423641|entityMap^0|2|5|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

It's the definition of the assignment operator: It might look better formatted like this (out-of-line definition):

<pre><code>X &amp; X::operator=(int z)
{
 temp = z;
 return *this;
}
</code></pre>

It allows you to write something like this:

<pre><code>X a, b, c(10);
a = b = c;
</code></pre>

The value of each assignment subexpression is a reference to the assignee.

The operator needs to be declared inside the class definition of course:

<pre><code>class X
{
public:
 X &amp; operator=(int);
 // ...
};
</code></pre>

I am reading a C++ book.
The program tries to make a vector of objects.
Here is the part I do not understand

<pre><code>class X {
public:
 X();
 X(int m) {
 temp = x;
 }

 int temp;


 X &amp;operator =(int z) {
 temp = z;
 return *this;
 }

private :
 // Some functions here
}
</code></pre>

What does above line mean?
Is this some sort of overloading?
Then how?

What does &operator mean in overloading

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在读一本C++的书。该程序试图创建一个对象的向量。这是我不明白的部分class X {public:    X();    X(int m) {        temp = x;    }    int temp;    X &operator =(int z) {        temp = z;        ...

问&运算符在重载中意味着什么？
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问&运算符在重载中意味着什么？EN