blocks|key|293436|text|发生这种情况的原因是与模板的查找规则有关。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|293437|p不是依赖表达式，因为它只是一个标识符，而不是依赖于模板参数的东西。这意味着不会搜索依赖于模板参数的基类来解析名称p。为了解决这个问题，你需要使用一些依赖于模板参数的东西。使用this->可以做到这一点。|offset|length|style|CODE|293438|例如：|293439|cout+<<+this->p+<<+endl;|code-block|syntax|javascript|293440|entityMap^0|0|0|1|1L|1|2G|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

The reason that this is happening is to do with the lookup rules for templates.

<code>p</code> isn't a dependent expression because it is just an identifier and not something that depends on the template parameter. This means that base classes that are dependent on the template parameter won't be searched to resolve the name <code>p</code>. To work around this issue you need to use something that does depend on the template parameter. Using <code>this-&gt;</code> will do this.

e.g.

<pre><code>cout &lt;&lt; this-&gt;p &lt;&lt; endl;
</code></pre>

blocks|key|296010|text|我在VC9中没有得到那个编译器错误。然而，代码有几个问题:首先，它不需要是一个模板类，因为它目前是written...but也许你只是为了这个问题简化了它？其次，基类应该有一个虚拟析构函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|296011|#include+<iostream>

using+namespace+std;

class+class0+{
public:
+++virtual+~class0(){}

protected:
++++char+p;
public:
++++char+getChar();
};

class+class1+:+public+class0+{
public:
++++void+printChar();
};

void+class1::printChar(){
++++cout+<<+p+<<+endl;//p+was+not+declared+in+this+scope
}

int+main()+{
+++class1+c;
+++c.printChar();
+++return+1;
}|code-block|syntax|javascript|296012|既然你正在学习模板，我建议你在学习的时候不要混淆概念(继承和模板)。从这样一个简单的例子开始...|296013|#include+<iostream>
#include+<string>

using+namespace+std;

template+<typename+T>
T+add(const+T&+a,+const+T&+b)+{
+++return+a+%2B+b;
}

int+main()+{
+++int+x+=+5;
+++int+y+=+5;

+++int+z+=+add(x,+y);
+++cout+<<+z+<<+endl;

+++string+s1("Hello,+");
+++string+s2("World!");

+++string+s3+=+add(s1,+s2);
+++cout+<<+s3+<<+endl;

+++return+1;
}|296014|上面代码中的重要概念是，我们编写了一个ONE函数，它知道如何添加整数和字符串(以及许多其他类型)。|offset|length|style|BOLD|296015|entityMap^0|0|0|0|0|J|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I don't get that compiler error in VC9. However, there are several problems with the code: First, it doesn't need to be a template class as it's currently written...but maybe you just simplified it for this question? Second, The base class should have a virtual destructor.

<pre><code>#include &lt;iostream&gt;

using namespace std;

class class0 {
public:
 virtual ~class0(){}

protected:
 char p;
public:
 char getChar();
};

class class1 : public class0 {
public:
 void printChar();
};

void class1::printChar(){
 cout &lt;&lt; p &lt;&lt; endl;//p was not declared in this scope
}

int main() {
 class1 c;
 c.printChar();
 return 1;
}
</code></pre>

Since you're learning about templates, I would suggest not mixing concepts (inheritance &amp; templates) while learning. Start with a simple example like this...

<pre><code>#include &lt;iostream&gt;
#include &lt;string&gt;

using namespace std;

template &lt;typename T&gt;
T add(const T&amp; a, const T&amp; b) {
 return a + b;
}

int main() {
 int x = 5;
 int y = 5;

 int z = add(x, y);
 cout &lt;&lt; z &lt;&lt; endl;

 string s1("Hello, ");
 string s2("World!");

 string s3 = add(s1, s2);
 cout &lt;&lt; s3 &lt;&lt; endl;

 return 1;
}
</code></pre>

The important concept in the code above is that we wrote ONE function that knows how to add integers and strings (and many other types for that matter).

blocks|key|3020235|text|要在依赖基类中查找名称，需要满足两个条件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3020236|名称依赖于unqualified|3020237|It's+，查找不是必需的，这是必要的|unordered-list-item|3020238|3020239|C%2B%2B03中声明的这些规则不同于rules+stated+by+unrevised+C%2B%2B98，后者满足第二个项目符号(使名称依赖)就足以查找在依赖基类中声明的名称。|offset|length|3020240|依赖名称是在实例化时查找的，而非非限定查找不会忽略依赖基类。要找到在依赖基类中声明的名称，需要满足这两个条件，单凭这两个条件是不够的。要同时满足这两个条件，可以使用各种构造|3020241|this->p
class1::p|code-block|syntax|javascript|3020242|两个名称p都是依赖的，第一个版本使用类成员访问查找，第二个版本使用限定名称查找。|style|CODE|3020243|entityMap|0|LINK|mutability|MUTABLE|url|http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#213^0|0|0|0|0|G|V|0|0|0|0|4|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@$J|19|K|1A|1|1B]]|A|$]]|$1|L|3|M|5|6|7|1C|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1D|8|@]|9|@]|A|$Q|R]]|$1|S|3|T|5|6|7|1E|8|@$J|1F|K|1G|U|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

For a name to be looked up in a dependent base class, two conditions need to be satisfied

<ul>
<li>It's necessary that the lookup is not unqualified</li>
<li>It's necessary that the name is dependent</li>
</ul>

These rules as stated in C++03 are different from the <a href="http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#213" rel="noreferrer">rules stated by unrevised C++98</a>, where satisfying the second bullet (making a name dependent) was sufficient for finding names declared in dependent base classes. 

A dependent name is looked up at instantiation time and a lookup other than unqualified lookup will not ignore dependent base classes. Both of these conditions need to be satisfied to find a name declared in a dependent base class, neither of them alone is sufficient. To satisfy both conditions you can use various constructs

<pre><code>this-&gt;p
class1::p
</code></pre>

Both names <code>p</code> are dependent and the first version uses class member access lookup and the second version uses qualified name lookup.

blocks|key|3020276|text|很抱歉重新提出了这样一个老问题，但我只是想添加这个东西，如果你的成员函数中有很多很多的"p“，我觉得它很有价值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3020277|class+class1:public+class0<T>+{
public:
++++using+class0<T>::p;+//+add+this+line+and+in+all+member+functions+
++++++++++++++++++++++++//+will+assume+that+"p"+is+the+p+from+class0<T>
++++++++++++++++++++++++//+very+useful+if+you+have+hundreds+of+"p":s
++++++++++++++++++++++++//+and+don't+want+to+replace+every+single+one+with+"this->p"
++++void+printChar();
};|code-block|syntax|javascript|3020278|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Sorry for reviving such an old question, but I just wanted to add this thing which i find valuable if you have lots and lots of "p" in your member functions.

<pre><code>class class1:public class0&lt;T&gt; {
public:
 using class0&lt;T&gt;::p; // add this line and in all member functions 
 // will assume that "p" is the p from class0&lt;T&gt;
 // very useful if you have hundreds of "p":s
 // and don't want to replace every single one with "this-&gt;p"
 void printChar();
};
</code></pre>

i am a new programmer in c++. and i am using templates for the first time.

i have an abstract class and another class extending it. but all the protected members of the abstract class are not recognised by the other class:

class0.h:

<pre><code>template&lt;class T&gt;
class class0 {

protected:
 char p;
public:
 char getChar();
};

**class1.h**
template&lt;class T&gt;
class class1:public class0&lt;T&gt; {
public:
 void printChar();
};
template&lt;class T&gt;
void class1&lt;T&gt;::printChar(){
 cout&lt;&lt; p&lt;&lt;endl;//p was not declared in this scope
}
</code></pre>

thank you. have a great week =)

template inheritance c++

编译器

我是c++的新程序员。我是第一次使用模板。我有一个抽象类和另一个扩展它的类。但抽象类的所有受保护成员不能被另一个类识别：类0.h：template<class T>class class0 {protected: char p;public: char getChar();};**class1.h**tem...

问模板继承c++
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