blocks|key|1340364|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1340365|错误运算符：‘C2801+=’必须是非静态成员|blockquote|offset|length|style|BOLD|1340366|1340367|1340368|粗体的词是这里的关键。friend不是会员，而是朋友。删除该friend关键字并将operator=视为成员：|CODE|1340369|语法上正确的版本是：|1340370|template+<class+T>
class+Matrice
{
++++T+m,n;
public:
++++template+<class+V>
++++Matrice<V>&+operator=(const+Matrice<V>+&);
};

template+<class+T>
template+<class+V>
Matrice<V>&+Matrice<T>::operator=(const+Matrice<V>+&M)
{
++++/*...*/
++++return+*this;
}|code-block|syntax|javascript|1340371|尽管我认为在那里使用该template+<class+V>是错误的；语义上正确的版本应该是|1340372|template+<class+T>
class+Matrice
{
++++T+m,n;
public:
++++Matrice<T>&+operator=(const+Matrice<T>+&);
};

template+<class+T>
Matrice<T>&+Matrice<T>::operator=(const+Matrice<T>+&M)
{
++++/*...*/
++++return+*this;
}|1340373|说明:您通常不希望以这种方式将+Type<V>分配给Type<T>；如果您必须这样做，那么这可能是糟糕设计的标志。|1340374|entityMap^0|0|L|2|0|0|0|B|6|U|6|15|9|0|0|0|B|I|0|0|G|7|Q|7|0^^$0|@$1|2|3|-4|4|5|6|11|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|12|7|@$D|13|E|14|F|G]]|8|@]|9|$]]|$1|H|3|-4|4|5|6|15|7|@]|8|@]|9|$]]|$1|I|3|-4|4|5|6|16|7|@]|8|@]|9|$]]|$1|J|3|K|4|5|6|17|7|@$D|18|E|19|F|L]|$D|1A|E|1B|F|L]|$D|1C|E|1D|F|L]]|8|@]|9|$]]|$1|M|3|N|4|5|6|1E|7|@]|8|@]|9|$]]|$1|O|3|P|4|Q|6|1F|7|@]|8|@]|9|$R|S]]|$1|T|3|U|4|5|6|1G|7|@$D|1H|E|1I|F|L]]|8|@]|9|$]]|$1|V|3|W|4|Q|6|1J|7|@]|8|@]|9|$R|S]]|$1|X|3|Y|4|5|6|1K|7|@$D|1L|E|1M|F|L]|$D|1N|E|1O|F|L]]|8|@]|9|$]]|$1|Z|3|-4|4|5|6|1P|7|@]|8|@]|9|$]]]|10|$]]

<blockquote>
 error C2801: 'operator =' must be a non-static member
</blockquote>

The bolded word is key here. <code>friend</code> is not a member; it's a friend. Remove that <code>friend</code> keyword and treat <code>operator=</code> as member:

The syntactically proper version is:

<pre><code>template &lt;class T&gt;
class Matrice
{
 T m,n;
public:
 template &lt;class V&gt;
 Matrice&lt;V&gt;&amp; operator=(const Matrice&lt;V&gt; &amp;);
};

template &lt;class T&gt;
template &lt;class V&gt;
Matrice&lt;V&gt;&amp; Matrice&lt;T&gt;::operator=(const Matrice&lt;V&gt; &amp;M)
{
 /*...*/
 return *this;
}
</code></pre>

Although I think that it's wrong to use that <code>template &lt;class V&gt;</code> there; the sematically proper version would be

<pre><code>template &lt;class T&gt;
class Matrice
{
 T m,n;
public:
 Matrice&lt;T&gt;&amp; operator=(const Matrice&lt;T&gt; &amp;);
};

template &lt;class T&gt;
Matrice&lt;T&gt;&amp; Matrice&lt;T&gt;::operator=(const Matrice&lt;T&gt; &amp;M)
{
 /*...*/
 return *this;
}
</code></pre>

Explanation: you don't generally want to assign <code>Type&lt;V&gt;</code> to <code>Type&lt;T&gt;</code> in this way; if you have to, then it is probably sign of bad design.

blocks|key|2976260|text|标准上说|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2976261|2976262|+12.8复制和移动类对象+class.copy|blockquote|offset|length|style|BOLD|2976263|2976264|..。|2976265|17用户声明的复制赋值运算符X::operator=是类X的非静态非模板成员函数，它只有一个参数，类型为X、X&、常量X&、易失性X&或常量易失性X&。|CODE|2976266|2976267|好友函数不能满足这些要求。它必须是成员函数。|2976268|2976269|为了解决这只是“普通”赋值，而不是复制赋值的评论，让我们从标准中添加另一个引用：|2976270|2976271|+13.5.3+Assignment+over.ass|2976272|2976273|赋值运算符应该由只有一个参数的非静态成员函数实现。|2976274|2976275|在标准语言中，“应当”不会留下任何其他选项。|2976276|entityMap^0|0|0|1|C|0|0|0|E|C|0|0|0|0|0|0|1|H|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|18|8|@$F|19|G|1A|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1D|8|@$F|1E|G|1F|H|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1H|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1J|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|W|3|X|5|E|7|1L|8|@$F|1M|G|1N|H|I]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1P|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|1R|8|@]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|15|$]]

The standard says
<blockquote>
12.8 Copying and moving class objects [class.copy]
...
17 A user-declared copy assignment operator <code>X::operator=</code> is a non-static non-template member function of class X with exactly one parameter of type X, X&amp;, const X&amp;, volatile X&amp; or const volatile X&amp;.
</blockquote>
A friend function doesn't fulfill these requirements. It must be a member function.
<hr />
To address the comments that this is just &quot;plain&quot; assignment, not copy assignment, let's add another quote from the standard:
<blockquote>
13.5.3 Assignment [over.ass]
An assignment operator shall be implemented by a non-static member function with exactly one parameter.
</blockquote>
In standardese, &quot;shall&quot; doesn't leave any options to do anything else.

blocks|key|1340450|text|您混合了朋友和成员的声明和定义:在第一个示例中，您将operator=声明为朋友，但将其定义为类成员。在第二个示例中，您将operator=声明为成员，但尝试将其定义为非成员。您的operator=必须是成员(请参阅this+question为什么)，并且您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1340451|template+<class+T>
class+Matrice
{
++++T+m,n;
public:
++++Matrice<T>&+operator=(Matrice<T>+&);
};

template+<class+T>
Matrice<T>&+Matrice<T>::operator=(Matrice<T>+&M)
{
++++/*...*/
++++return+*this;
}|code-block|syntax|javascript|1340452|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3933637/why-friend-function-cant-be-used-for-overloading-assignment-operator^0|Q|9|1P|9|2I|9|30|D|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You mixed friend and member declarations and definitions: in the first example you declared <code>operator=</code> as a friend but defined it as a class member. In the second example, you declared <code>operator=</code> as a member but tried to define it as a non-member. Your <code>operator=</code> has to be a member (see <a href="https://stackoverflow.com/questions/3933637/why-friend-function-cant-be-used-for-overloading-assignment-operator">this question</a> why) and you can do the following:

<pre><code>template &lt;class T&gt;
class Matrice
{
 T m,n;
public:
 Matrice&lt;T&gt;&amp; operator=(Matrice&lt;T&gt; &amp;);
};

template &lt;class T&gt;
Matrice&lt;T&gt;&amp; Matrice&lt;T&gt;::operator=(Matrice&lt;T&gt; &amp;M)
{
 /*...*/
 return *this;
}
</code></pre>

I am trying to overload the <code>operator=</code> in a template class.

I have this template class:

<pre><code>template &lt;class T&gt;
class Matrice
{
 T m,n;
public:
 template &lt;class V&gt;
 friend Matrice&lt;V&gt;&amp; operator=(const Matrice&lt;V&gt; &amp;);
};

template &lt;class T&gt;
Matrice&lt;T&gt;&amp; Matrice&lt;T&gt;::operator=(const Matrice&lt;T&gt; &amp;M)
{
 /*...*/
 return *this;
}
</code></pre>

and I also tried:

<pre><code>template &lt;class T&gt;
class Matrice
{
 T m,n;
public:
 template &lt;class V&gt;
 Matrice&lt;V&gt;&amp; operator=(Matrice&lt;V&gt; &amp;);
};

template &lt;class T&gt;
Matrice&lt;T&gt;&amp; operator=(Matrice&lt;T&gt; &amp;M)
{
 /*...*/
 return *this;
}
</code></pre>

but I still get this error:

<pre><code>error C2801: 'operator =' must be a non-static member
</code></pre>

C++ error C2801: 'operator =' must be a non-static member

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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我正在尝试在模板类中重载operator=。我有这个模板类：template <class T>class Matrice{ T m,n;public: template <class V> friend Matrice<V>& operator=(const Matrice<V> &);};temp...

问C++错误运算符：‘C2801 =’必须是非静态成员
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问C++错误运算符：‘C2801 =’必须是非静态成员EN