blocks|key|890760|text|这是一个有点奇怪的初始化字符串的方式，如果你事先知道你想要在其中存储什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|890761|您可以简单地使用：|890762|String+test="c";|code-block|syntax|javascript|890763|如果您有特殊需要将字符变量转换为字符串，可以使用内置的ToString()函数：|offset|length|style|CODE|890764|String+test+=+myCharVariable.ToString();|890765|entityMap^0|0|0|0|R|A|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

That is a bit of a strange way to initialize a string, if you know beforehand what you want to store in it.

You can simply use:

<pre><code>String test="c";
</code></pre>

If you have a specific need to convert a char variable to a string, you can use the built in <code>ToString()</code> function:

<pre><code>String test = myCharVariable.ToString();
</code></pre>

blocks|key|2299245|text|String类有许多构造函数，如果您要做的只是创建一个包含一个字符的字符串，您可以使用以下方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2299246|String+test+=+new+String(new+char[]+{+'c'+});|code-block|syntax|javascript|2299247|entityMap^0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The <code>String</code> class has many constructors, if all you're after is to create a string containing one character, you can use the following:

<pre><code>String test = new String(new char[] { 'c' });
</code></pre>

blocks|key|890850|text|string+myString1+=+new+string(new+char[]+{'a'});
string+myString2+=+'a'.ToString();
string+myString3+=+"a";
string+myString4+=+new+string('a',+1);
unsafe+{
++++char+a+=+'a';
++++string+myString5+=+new+string(&a);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|890851|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>string myString1 = new string(new char[] {'a'});
string myString2 = 'a'.ToString();
string myString3 = "a";
string myString4 = new string('a', 1);
unsafe {
 char a = 'a';
 string myString5 = new string(&amp;a);
}
</code></pre>

blocks|key|2299261|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2299262|String+test("Something");|code-block|syntax|javascript|2299263|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|$]]

use 

<pre><code>String test("Something");
</code></pre>

blocks|key|711118|text|unsafe
{
++++char+c+=+'c';
++++char+*ch+=+&c;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|711119|你的例子有一个String和一个编译错误，因为使用了一个字符串构造函数重载，所以我猜你真的只想要一个字符数组，也就是一个String，也许不是一个char*。|unstyled|offset|length|style|CODE|711120|在这种情况下：|711121|char+c+=+'c';
string+s+=+c.ToString();+//+or...
string+s1+=+""+%2Bc;|711122|还可以使用：|711123|unsafe
{
++++char+c+=+'c';
++++char*+ch+=+&c;
++++string+s1+=+new+string(ch);
++++string+s2+=+new+string(c,+0);
}|711124|entityMap^0|0|7|6|1O|6|21|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@$G|W|H|X|I|J]|$G|Y|H|Z|I|J]|$G|10|H|11|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$B|C]]|$1|O|3|P|5|F|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$B|C]]|$1|S|3|-4|5|F|7|16|8|@]|9|@]|A|$]]]|T|$]]

<pre><code>unsafe
{
 char c = 'c';
 char *ch = &amp;c;
}
</code></pre>

Your example has a <code>String</code> and a compile error from using one of the String constructor overloads, so I'm guessing you really just want an array of chars, aka a <code>String</code> and maybe not a <code>char*</code>.

In which case:

<pre><code>char c = 'c';
string s = c.ToString(); // or...
string s1 = "" +c;
</code></pre>

Also available:

<pre><code>unsafe
{
 char c = 'c';
 char* ch = &amp;c;
 string s1 = new string(ch);
 string s2 = new string(c, 0);
}
</code></pre>

blocks|key|2299359|text|字符串测试=新字符串(新char[]{'c'})；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2299360|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

String test = new String(new char[]{'c'});

blocks|key|2299403|text|如果你在硬编码，有没有什么原因不能直接使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2299404|String+test+=+"c";|code-block|syntax|javascript|2299405|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|$]]

If you are hard coding it, is there a reason you cant just use:

<pre><code>String test = "c";
</code></pre>

blocks|key|2299413|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2299414|var+test+=+'c'.ToString()|code-block|syntax|javascript|2299415|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|$]]

How about:

<pre><code>var test = 'c'.ToString()
</code></pre>

blocks|key|891094|text|从您的示例中执行此转换的最简单方法是更改您正在使用的单引号的类型|type|unstyled|depth|inlineStyleRanges|entityRanges|data|891095|String+test+=+new+String('c');|code-block|syntax|javascript|891096|要使用双引号并删除构造函数调用，请执行以下操作：|891097|String+test+=+"c";|891098|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The easiest way to do this conversion from your example is just change the type of quotes you are using from single quotes

<pre><code>String test = new String('c');
</code></pre>

to double quotes and remove the constructor call:

<pre><code>String test = "c";
</code></pre>

blocks|key|2014915|text|有另一个答案也无伤大雅：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2014916|string+test+=+string.Empty+%2B+'c';|code-block|syntax|javascript|2014917|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|$]]

Can't hurt to have yet another answer:

<pre><code>string test = string.Empty + 'c';
</code></pre>

blocks|key|711892|text|在String构造函数中使用char时，还应提供count参数来指定应将该字符添加到字符串中的次数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|711893|String+test=new+String('c',+1);|code-block|syntax|javascript|711894|另请参见here。|offset|length|711895|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/xsa4321w.aspx^0|0|0|4|4|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

When using a char in the String constructor, you should also give a count parameter to specify how many times that character should be added to the string:

<pre><code>String test=new String('c', 1);
</code></pre>

See also <a href="http://msdn.microsoft.com/en-us/library/xsa4321w.aspx" rel="nofollow noreferrer">here</a>.

blocks|key|2011888|text|char+c+=+'R';
char+*pc+=+&c;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2011889|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|$]]

blocks|key|711908|text|对于接受单个char作为参数的String，不存在公共构造函数的重载。最接近的匹配是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|711909|public+String(char+c,+int+count)|code-block|syntax|javascript|711910|它创建了一个重复char+c+count时间的新String。因此，您可以这样说|711911|string+s+=+new+string('c',+1);|711912|还有其他选择。有一个接受char[]作为参数的String的公共构造函数：|711913|public+String(char[]+value)|711914|这将创建一个用value中的Unicode字符初始化的String。因此你可以说|711915|char+c+=+'c';
string+s+=+new+String(new+char[]+{+c+});|711916|另一种选择是说|711917|char+c+=+'c'
string+s+=+c.ToString();|711918|但大多数人期望看到的最直接的方法是|711919|string+s+=+"c";|711920|至于将char转换为char+*，您不能安全地这样做。如果要使用接受char+*作为参数的String的公共构造函数的重载，可以这样做：|711921|unsafe+{
++++char+c+=+'c';
++++char+*p+=+&c;
++++string+s+=+new+string(p);
}|711922|entityMap^0|6|4|F|6|0|0|8|C|O|6|0|0|C|6|N|6|0|0|7|5|R|6|0|0|0|0|0|0|3|4|A|6|Y|6|19|6|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1R|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1Z|8|@]|D|@]|E|$I|J]]|$1|10|3|11|5|6|7|20|8|@]|D|@]|E|$]]|$1|12|3|13|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|14|3|15|5|6|7|22|8|@$9|23|A|24|B|C]|$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]]|D|@]|E|$]]|$1|16|3|17|5|H|7|2B|8|@]|D|@]|E|$I|J]]|$1|18|3|-4|5|6|7|2C|8|@]|D|@]|E|$]]]|19|$]]

There is no overload of the public constructor for <code>String</code> that accepts a single <code>char</code> as a parameter. The closest match is

<pre><code>public String(char c, int count)
</code></pre>

which creates a new <code>String</code> that repeats the <code>char c count</code> times. Thus, you could say

<pre><code>string s = new string('c', 1);
</code></pre>

There are other options. There is a public constructor of <code>String</code> that accepts a <code>char[]</code> as a parameter:

<pre><code>public String(char[] value)
</code></pre>

This will create a <code>String</code> that is initialized with the Unicode characters in <code>value</code>. Thus you could say

<pre><code>char c = 'c';
string s = new String(new char[] { c });
</code></pre>

Another option is to say

<pre><code>char c = 'c'
string s = c.ToString();
</code></pre>

But the most straightforward approach that most will expect to see is

<pre><code>string s = "c";
</code></pre>

As for converting a <code>char</code> to a <code>char *</code> you can not safely do this. If you want to use the overload of the public constructor for <code>String</code> that accepts a <code>char *</code> as a parameter, you could do this:

<pre><code>unsafe {
 char c = 'c';
 char *p = &amp;c;
 string s = new string(p);
}
</code></pre>

blocks|key|2011921|text|使用单引号(如您的问题：'c')意味着您正在创建一个char。使用双引号，例如"c"，意味着您创建了一个string。这些在c#中是不可互换的类型。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2011922|正如您可能知道的，char*在某种程度上是字符串在c%2B%2B中的表示方式，并且c#支持c%2B%2B的一些约定。这意味着在c#中可以很容易地(至少对程序员而言)将char*转换为string。不幸的是，char本身并不是char*，因此也无法做到这一点。|2011923|entityMap^0|C|3|Q|4|13|3|1G|6|0|9|5|23|5|2B|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|I|$]]

Using single quotes (as in your question: <code>'c'</code>) means that you are creating a <code>char</code>. Using double quotes, e.g. <code>"c"</code>, means you creating a <code>string</code>. These are not interchangable types in c#.

A <code>char*</code>, as you might be aware, is how strings are represented in c++ to some extent, and c# supports some of the conventions of c++. This means that a <code>char*</code> can easily (for the programmer at least) be converted to a <code>string</code> in c#. Unfortunately a char is not inherently a char*, so the same cannot be done.

how can I convert a char to a char* in c#?

I'm initializeing a String object like this:

<pre><code>String test=new String('c');
</code></pre>

and I'm getting this error:

<code>Argument '1': cannot convert from 'char' to 'char*'</code>

how can I convert a char to a char* 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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如何在c#中将字符转换为字符*？我正在初始化一个字符串对象，如下所示：String test=new String('c');我得到了这个错误：Argument '1': cannot convert from 'char' to 'char*'

问如何在c#中将字符转换为字符*？
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问如何在c#中将字符转换为字符*？EN