blocks|key|1529736|text|你不能，至少不能直接这么做。将一个或一系列标记转换为字符串文字的唯一方法是在宏内使用预处理器的字符串化运算符(#)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1529737|如果想要获得表示类型的字符串文字，则必须自己编写一些内容，可能是使用宏来实例化模板并将字符串类型名称传递给它。|1529738|任何通用方法的一个问题是:应该为以下用途提供什么字符串：|1529739|Matrix<char>+x;
typedef+char+MyChar;
Matrix<MyChar>+y;|code-block|syntax|javascript|1529740|x和y都属于同一类型，但一个直接使用char，另一个使用类型定义MyChar。|1529741|entityMap^0|1J|1|0|0|0|0|0|1|2|1|I|4|W|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

You can't, at least not directly. The only way to convert a token or series of tokens into a string literal is using the preprocessor's stringization operator (<code>#</code>) inside of a macro. 

If you want to get a string literal representing the type, you'll have to write something yourself, perhaps by using a macro to instantiate the template and pass it the stringized type name.

One problem with any general approach is: what string should be given for the following uses:

<pre><code>Matrix&lt;char&gt; x;
typedef char MyChar;
Matrix&lt;MyChar&gt; y;
</code></pre>

Both <code>x</code> and <code>y</code> are of the same type, but one uses <code>char</code> directly and the other uses the typedef <code>MyChar</code>.

blocks|key|4020586|text|如果类型是基类型之一，则在string中获取类型名称是不可能的。对于用户定义的类型，可以使用typeid(my_type).name()。您还需要#include+<typeinfo>+:)+more+info...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4020587|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Typeid^0|D|6|1A|M|21|J|2O|C|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

It is impossilbe to get name of type in <code>string</code> if the type is one of base types. For user defined types you can use <code>typeid(my_type).name()</code>. Also you need <code>#include &lt;typeinfo&gt;</code> :)
<a href="http://en.wikipedia.org/wiki/Typeid" rel="noreferrer">more info...</a>

blocks|key|2424317|text|这方面没有内置的机制。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2424318|typeid(T)::name()可以提供一些信息，但标准并没有要求这个字符串必须是人类可读的；只是对于每种类型，它必须是不同的。(E.x。(+Microsoft+Visual+C%2B%2B使用人类可读的字符串；而GCC不使用。)|offset|length|style|CODE|2424319|不过，您可以构建自己的系统。例如，基于特征的。如下所示：|2424320|//+default+implementation
template+<typename+T>
struct+TypeName
{
++++static+const+char*+Get()
++++{
++++++++return+typeid(T).name();
++++}
};

//+a+specialization+for+each+type+of+those+you+want+to+support
//+and+don't+like+the+string+returned+by+typeid
template+<>
struct+TypeName<int>
{
++++static+const+char*+Get()
++++{
++++++++return+"int";
++++}
};

//+usage:
const+char*+name+=+TypeName<MyType>::Get();|code-block|syntax|javascript|2424321|entityMap^0|0|0|H|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]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

There is no built-in mechanism for this.
<code>typeid(T)::name()</code> can give some info, but the standard does not mandate this string to be human-readable; just that it has to be distinct for each type. (E.x. Microsoft Visual C++ uses human-readable strings; GCC does not.)
You can build your own system though. For example, traits-based. Something like this:
<pre><code>// default implementation
template &lt;typename T&gt;
struct TypeName
{
 static const char* Get()
 {
 return typeid(T).name();
 }
};

// a specialization for each type of those you want to support
// and don't like the string returned by typeid
template &lt;&gt;
struct TypeName&lt;int&gt;
{
 static const char* Get()
 {
 return &quot;int&quot;;
 }
};

// usage:
const char* name = TypeName&lt;MyType&gt;::Get();
</code></pre>

blocks|key|1527384|text|对GCC来说，你必须使用一个技巧。使用cxxabi.h，我为此编写了一个小包装器：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1527385|#include+<string>
#include+<iostream>
#include+<iomanip>
#include+<typeinfo>
#include+<cxxabi.h>

#define+DEBUG_TYPE(x)+do+{+typedef+void(*T)x;+debug_type<T>(T(),+#x);+}+while(0)

template<typename+T>
struct+debug_type
{
++++template<typename+U>
++++debug_type(void(*)(U),+const+std::string&+p_str)
++++{
++++++++std::string+str(p_str.begin()+%2B+1,+p_str.end()+-+1);
++++++++std::cout+<<+str+<<+"+=>+";
++++++++char+*+name+=+0;
++++++++int+status;
++++++++name+=+abi::__cxa_demangle(typeid(U).name(),+0,+0,+&status);
++++++++if+(name+!=+0)+{+std::cout+<<+name+<<+std::endl;+}
++++++++else+{+std::cout+<<+typeid(U).name()+<<+std::endl;+}
++++++++free(name);
++++}
};|code-block|syntax|javascript|1527386|双括号是必要的。适用于任何类型。|1527387|现在，您可以将其用于boost::mpl|1527388|DEBUG_TYPE((if_c<true,+true_,+false_>::type));|1527389|将打印：|1527390|if_c<true,+true_,+false_>::type+=>+bool_<true>|1527391|entityMap^0|J|8|0|0|0|A|A|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|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|V|$]]

For GCC you have to use a trick. Using <code>cxxabi.h</code>, I wrote a little wrapper for this purpose:
<pre><code>#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;iomanip&gt;
#include &lt;typeinfo&gt;
#include &lt;cxxabi.h&gt;

#define DEBUG_TYPE(x) do { typedef void(*T)x; debug_type&lt;T&gt;(T(), #x); } while(0)

template&lt;typename T&gt;
struct debug_type
{
 template&lt;typename U&gt;
 debug_type(void(*)(U), const std::string&amp; p_str)
 {
 std::string str(p_str.begin() + 1, p_str.end() - 1);
 std::cout &lt;&lt; str &lt;&lt; &quot; =&gt; &quot;;
 char * name = 0;
 int status;
 name = abi::__cxa_demangle(typeid(U).name(), 0, 0, &amp;status);
 if (name != 0) { std::cout &lt;&lt; name &lt;&lt; std::endl; }
 else { std::cout &lt;&lt; typeid(U).name() &lt;&lt; std::endl; }
 free(name);
 }
};
</code></pre>
The double parentheses are necessary. Will work with any type.
Now you can use it for <code>boost::mpl</code>:
<pre><code>DEBUG_TYPE((if_c&lt;true, true_, false_&gt;::type));
</code></pre>
will print:
<pre><code>if_c&lt;true, true_, false_&gt;::type =&gt; bool_&lt;true&gt;
</code></pre>

blocks|key|2424369|text|变通方法...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2424370|#define+Tprint(x)+print<x>(#x)

template<typename+T>
void+print+(string+ltype){
cout<<ltype<<"+=+"<<sizeof(T)<<endl;
}|code-block|syntax|javascript|2424371|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|$]]

workaround way...

<pre><code>#define Tprint(x) print&lt;x&gt;(#x)

template&lt;typename T&gt;
void print (string ltype){
cout&lt;&lt;ltype&lt;&lt;" = "&lt;&lt;sizeof(T)&lt;&lt;endl;
}
</code></pre>

blocks|key|356589|text|你可以使用C%2B%2B+reflection+library。所以：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|356590|using+namespace+ponder;

Class::declare<Matrix>();

std::string+const&+name+=+classByType<Matrix>().name();|code-block|syntax|javascript|356591|一旦您有了元类信息，这也为您提供了其他选项，比如查看类成员是什么。|356592|entityMap|0|LINK|mutability|MUTABLE|url|http://billyquith.github.io/ponder/^0|5|M|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

You could use a <a href="http://billyquith.github.io/ponder/" rel="nofollow">C++ reflection library</a>. So:

<pre><code>using namespace ponder;

Class::declare&lt;Matrix&gt;();

std::string const&amp; name = classByType&lt;Matrix&gt;().name();
</code></pre>

This gives you other options as well once you have the metaclass information, like looking what the class members are.

blocks|key|4020704|text|template<+typename+From,typename+To>
static+inline+bool+superConvert(const+From&+fromVar,To&+toVar)
{
++++stringstream+ss;
++++ss<<fromVar;
++++ss>>toVar;
++++if(ss.fail())
++++{
++++++++return+false;
++++}
++++else
++++{
++++++++From+tempFrom;
++++++++stringstream+ss;
++++++++ss<<toVar;
++++++++ss>>tempFrom;
++++++++if(tempFrom+!=+fromVar)
++++++++{
++++++++++++return+false;
++++++++}
++++++++else
++++++++{
++++++++++++return+true;
++++++++}
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4020705|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>template&lt; typename From,typename To&gt;
static inline bool superConvert(const From&amp; fromVar,To&amp; toVar)
{
 stringstream ss;
 ss&lt;&lt;fromVar;
 ss&gt;&gt;toVar;
 if(ss.fail())
 {
 return false;
 }
 else
 {
 From tempFrom;
 stringstream ss;
 ss&lt;&lt;toVar;
 ss&gt;&gt;tempFrom;
 if(tempFrom != fromVar)
 {
 return false;
 }
 else
 {
 return true;
 }
 }
}
</code></pre>

While playing with templates in c++ I encountered a problem converting typename T to string. For example:

<pre><code>template &lt;typename T&gt;
class Matrix {
 public:
 Matrix() {
 //my_type = string type of T. i.e. if T is char. I want my_type to be "char".
 }
 string my_type;
}
</code></pre>

How do I convert T to a string that says what T is.

Note: I'm just playing around so please do not worry about when one might need such a thing.

How to convert typename T to string in c++

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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在c++中使用模板时，我遇到了将typename转换为字符串的问题。例如：template <typename T>class Matrix { public: Matrix() { //my_type = string type of T. i.e. if T is char. ...

问如何在c++中将typename转换为字符串
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问如何在c++中将typename转换为字符串EN