blocks|key|3768985|text|Process+<Sample+<double>+*>+a;
a.process(s);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3768986|上面代码的Process.process签名应该是void+process(Sample+<double>+**)，因为Item是Sample+<double>+*+-您希望它是void+process(Sample+<double>+*)|unstyled|offset|length|style|CODE|3768987|更改此设置：|3768988|Process+<Sample+<double>+*>+a;|3768989|至：|3768990|Process+<Sample+<double>+>+a;|3768991|如果您查看完整的错误消息，编译器很可能会告诉您在尝试将一个T与另一个Sample<T>进行匹配时使用了两个不同的Sample<T>|3768992|entityMap^0|0|5|F|P|W|1O|4|1T|H|2H|V|0|0|0|0|0|T|1|Y|9|1K|9|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]|$G|10|H|11|I|J]|$G|12|H|13|I|J]|$G|14|H|15|I|J]|$G|16|H|17|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|S|3|T|5|F|7|1C|8|@$G|1D|H|1E|I|J]|$G|1F|H|1G|I|J]|$G|1H|H|1I|I|J]]|9|@]|A|$]]|$1|U|3|-4|5|F|7|1J|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>Process &lt;Sample &lt;double&gt; *&gt; a;
a.process(s);
</code></pre>

The <code>Process.process</code> signature for the above code would be <code>void process(Sample &lt;double&gt; **)</code> since <code>Item</code> is <code>Sample &lt;double&gt; *</code> - You want it to be <code>void process(Sample &lt;double&gt; *)</code>

Change this:

<pre><code>Process &lt;Sample &lt;double&gt; *&gt; a;
</code></pre>

To:

<pre><code>Process &lt;Sample &lt;double&gt; &gt; a;
</code></pre>

If you look at the full error message, likely the compiler is telling you about the two different <code>T</code>'s used when trying to match one <code>Sample&lt;T&gt;</code> with another <code>Sample&lt;T&gt;</code>

blocks|key|2198993|text|Process+<Item>::process()需要一个Item*，所以Process+<Sample+<double>+*>::process()需要一个Sample+<double>**。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2198994|然而，解决方案要简单得多：|2198995|Sample+<double>+*s+=+new+Sample+<double>();
Process+<Sample+<double>+>+a;
a.process(s);|code-block|syntax|javascript|2198996|对于Process来说，对象是如何分配的一点都不重要，它得到的只是一个指向它的指针并使用它(假设它不会保持所有权或试图删除该对象)。|2198997|entityMap^0|0|P|T|5|11|12|27|H|0|0|0|2|7|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

<code>Process &lt;Item&gt;::process()</code> expects a <code>Item*</code>, so <code>Process &lt;Sample &lt;double&gt; *&gt;::process()</code> expects a <code>Sample &lt;double&gt;**</code>.

The solution is much easier, however:

<pre><code>Sample &lt;double&gt; *s = new Sample &lt;double&gt;();
Process &lt;Sample &lt;double&gt; &gt; a;
a.process(s);
</code></pre>

For <code>Process</code>, it doesn't matter at all how the object was allocated, all it gets is a pointer to it and works with it (assuming it won't maintain ownership or attempt to delete the object).

blocks|key|2199005|text|Sample+<double>+*s+=+new+Sample+<double>();
Process+<Sample+<double>+*>+a;+//<<-----------+here+you+go+wrong!
a.process(s);+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2199006|Process的参数应该仍然是Sample<double>，而不是Sample<double>*，因为后者会生成Item*+=>+Sample<double>**，而您在调用process成员函数时没有传递它。|unstyled|offset|length|style|CODE|2199007|所以正确的代码应该是：|2199008|Process+<Sample<double>+>+a;+//<<-----------+now+it's+correct!
a.process(s);|2199009|现在应该可以用了！|2199010|或者执行以下操作：|2199011|Sample+<double>+*s+=+new+Sample+<double>();
Process+<Sample+<double>+*>+a;+//<------+your+version!
a.process(+&s+);+++++++++++++++++//<------+note+I'm+passing+pointer+to+pointer!
+++++++//+%5E%5E+note+the+ampersand+(&)|2199012|entityMap^0|0|0|7|F|E|X|F|1K|5|1T|G|2F|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]|$G|10|H|11|I|J]|$G|12|H|13|I|J]|$G|14|H|15|I|J]|$G|16|H|17|I|J]|$G|18|H|19|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|1A|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|1C|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|1D|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1E|8|@]|9|@]|A|$B|C]]|$1|U|3|-4|5|F|7|1F|8|@]|9|@]|A|$]]]|V|$]]

<pre><code>Sample &lt;double&gt; *s = new Sample &lt;double&gt;();
Process &lt;Sample &lt;double&gt; *&gt; a; //&lt;&lt;----------- here you go wrong!
a.process(s); 
</code></pre>

The argument to <code>Process</code> should still be <code>Sample&lt;double&gt;</code>, not <code>Sample&lt;double&gt;*</code>, since the latter makes <code>Item*</code> => <code>Sample&lt;double&gt;**</code>, which you're not passing when calling <code>process</code> member function. 

So the correct code should be this:

<pre><code>Process &lt;Sample&lt;double&gt; &gt; a; //&lt;&lt;----------- now it's correct!
a.process(s);
</code></pre>

This should work now!

Or do this:

<pre><code>Sample &lt;double&gt; *s = new Sample &lt;double&gt;();
Process &lt;Sample &lt;double&gt; *&gt; a; //&lt;------ your version!
a.process( &amp;s ); //&lt;------ note I'm passing pointer to pointer!
 // ^^ note the ampersand (&amp;)
</code></pre>

blocks|key|3769091|text|问题不在于对象是以某种方式分配的。不可能判断一个对象是在堆栈上还是在空闲存储上分配的，至少通过任何可移植的、标准的方式。对于“栈”对象和“空闲存储”对象，通过指针访问对象成员的方式是完全相同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3769092|实际问题是您向第二个代码片段中的Process<>变量定义传递了错误的类型：|offset|length|style|CODE|3769093|Process+<Sample+<double>+*>+a;|code-block|syntax|javascript|3769094|看一看Process<>的定义|3769095|template+<typename+Item>
class+Process
{
public:
++++void+process+(Item+*item)
++++{+/*+...+*/+}
};|3769096|当Item的类型为Sample+<double>+*时，process()的函数签名为：|3769097|void+process+(Sample+<double>**+item)
{+/*+...+*/+}|3769098|显然这不是你想要的。|3769099|要修复此问题，请更改|3769100|Process+<Sample+<double>+*>+a;|3769101|至|3769102|Process+<Sample+<double>+>+a;|3769103|使用后者，process()的函数签名就变成了void+process+(Sample+<double>*+item)，这应该允许编译您的两个代码片段。|3769104|entityMap^0|0|G|9|0|0|3|9|0|0|1|4|9|H|S|9|0|0|0|0|0|0|0|5|9|N|10|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@$D|1A|E|1B|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1D|8|@$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1H|8|@$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1O|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1P|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|J|7|1R|8|@]|9|@]|A|$K|L]]|$1|10|3|11|5|6|7|1S|8|@]|9|@]|A|$]]|$1|12|3|13|5|J|7|1T|8|@]|9|@]|A|$K|L]]|$1|14|3|15|5|6|7|1U|8|@$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|17|$]]

The issue isn't the fact that the object was allocated a certain way. It's not possible to tell whether if an object was allocated on the stack or the free store given its pointer alone, at least though any portable, standard means. Accessing members of an object through a pointer happens the exact same way for both "stack" objects and "free store" objects.

The actual issue is that you're passing the wrong type to the <code>Process&lt;&gt;</code> variable definition in the second snippet:

<pre><code>Process &lt;Sample &lt;double&gt; *&gt; a;
</code></pre>

Take a look at the definition of <code>Process&lt;&gt;</code>:

<pre><code>template &lt;typename Item&gt;
class Process
{
public:
 void process (Item *item)
 { /* ... */ }
};
</code></pre>

When <code>Item</code> is of type <code>Sample &lt;double&gt; *</code>, then the function signature for <code>process()</code> becomes:

<pre><code>void process (Sample &lt;double&gt;** item)
{ /* ... */ }
</code></pre>

Which is not what you want, apparently.

To fix this, change

<pre><code>Process &lt;Sample &lt;double&gt; *&gt; a;
</code></pre>

to 

<pre><code>Process &lt;Sample &lt;double&gt; &gt; a;
</code></pre>

With the latter, the function signature for <code>process()</code> becomes <code>void process (Sample &lt;double&gt;* item)</code>, which should allow both of your code snippets to compile.

What is wrong in this code? 

<pre><code>template &lt;typename T&gt;
class Sample
{
public:
 T first;
 T second;
 typedef T Type;
};

and 

template &lt;typename Item&gt; 
class Process
{
public:
 void process (Item *item) 
 {
 typename Item::Type var = item-&gt;first + item-&gt;second;
 //Some code...
 }
};
</code></pre>

The method "process" should be able to work with an object allocated both "static" and "dynamic"...The first option works

<pre><code>int main(int argc, _TCHAR* argv[])
{
Sample &lt;double&gt; s;
Process &lt;Sample &lt;double&gt; &gt; a;
a.process(&amp;s);

return 0;
}
</code></pre>

but the second not

<pre><code>int main(int argc, _TCHAR* argv[])
{

Sample &lt;double&gt; *s = new Sample &lt;double&gt;();
Process &lt;Sample &lt;double&gt; *&gt; a;
a.process(s); //Error C2664: 'Process&lt;Item&gt;::process' : cannot convert parameter 1 from 'Sample&lt;T&gt; *' to 'Sample&lt;T&gt; *'
return 0;
}
</code></pre>

How to design a class and method "process" so as to be able to work with an object allocated both "static" and "dynamic" ? Thanks for your help...

C++ templates, problem with object allocated both "static" and "dynamic"

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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这段代码中有什么错误？template <typename T>class Sample{public: T first; T second; typedef T Type;};and template <typename Item> class Process{public: void pro...

问C++模板，分配“静态”和“动态”对象的问题
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问C++模板，分配“静态”和“动态”对象的问题EN