blocks|key|3423632|text|您可以利用数字的字符编码都是从48+(对于'0')到57+(对于'9')的顺序。这适用于ASCII、UTF-x和几乎所有其他编码(有关这方面的更多信息，请参阅下面的注释)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423633|因此，任何数字的整数值都是数字减去'0‘(或48)。|3423634|char+c+=+'1';
int+i+=+c+-+'0';+//+i+is+now+equal+to+1,+not+'1'|code-block|syntax|javascript|3423635|的同义词|3423636|char+c+=+'1';
int+i+=+c+-+48;+//+i+is+now+equal+to+1,+not+'1'|3423637|但是，我发现第一个c+-+'0'的可读性要好得多。|offset|length|style|CODE|3423638|entityMap^0|0|0|0|0|0|9|7|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@$O|10|P|11|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You can utilize the fact that the character encodings for digits are all in order from 48 (for '0') to 57 (for '9'). This holds true for ASCII, UTF-x and practically all other encodings (see comments below for more on this).

Therefore the integer value for any digit is the digit minus '0' (or 48).

<pre><code>char c = '1';
int i = c - '0'; // i is now equal to 1, not '1'
</code></pre>

is synonymous to

<pre><code>char c = '1';
int i = c - 48; // i is now equal to 1, not '1'
</code></pre>

However I find the first <code>c - '0'</code> far more readable.

blocks|key|206148|text|如果您担心编码问题，您可以随时使用Switch语句。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206149|只是要注意你保存这些大数字的格式。在某些系统中，整数的最大大小低至65,535+(32,767有符号)。其他系统，您有2,147,483,647+(或4,294,967,295未签名)|206150|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

If you are worried about encoding, you can always use a switch statement. 

Just be careful with the format you keep those large numbers in. The maximum size for an integer in some systems is as low as 65,535 (32,767 signed). Other systems, you've got 2,147,483,647 (or 4,294,967,295 unsigned)

blocks|key|3423681|text|或者，您可以使用“正确”方法，类似于最初的atoi方法，但使用std::stringstream。这应该可以使用字符作为输入，也可以使用字符串。(boost::lexical_cast是另一个更方便的语法选项)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423682|(atoi是一个旧的C函数，通常建议在可能的情况下使用更灵活和类型安全的C%2B%2B等效项。std::stringstream涵盖了字符串之间的转换)|3423683|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Or you could use the "correct" method, similar to your original atoi approach, but with std::stringstream instead. That should work with chars as input as well as strings. (boost::lexical_cast is another option for a more convenient syntax)

(atoi is an old C function, and it's generally recommended to use the more flexible and typesafe C++ equivalents where possible. std::stringstream covers conversion to and from strings)

blocks|key|3423708|text|下面的方法有什么问题吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3423709|int+CharToInt(const+char+c)
{
++++switch+(c)
++++{
++++case+'0':
++++++++return+0;
++++case+'1':
++++++++return+1;
++++case+'2':
++++++++return+2;
++++case+'3':
++++++++return+3;
++++case+'4':
++++++++return+4;
++++case+'5':
++++++++return+5;
++++case+'6':
++++++++return+6;
++++case+'7':
++++++++return+7;
++++case+'8':
++++++++return+8;
++++case+'9':
++++++++return+9;
++++default:
++++++++return+0;
++++}
}|code-block|syntax|javascript|3423710|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|$]]

Any problems with the following way of doing it?

<pre><code>int CharToInt(const char c)
{
 switch (c)
 {
 case '0':
 return 0;
 case '1':
 return 1;
 case '2':
 return 2;
 case '3':
 return 3;
 case '4':
 return 4;
 case '5':
 return 5;
 case '6':
 return 6;
 case '7':
 return 7;
 case '8':
 return 8;
 case '9':
 return 9;
 default:
 return 0;
 }
}
</code></pre>

blocks|key|206200|text|#include<iostream>
#include<stdlib>
using+namespace+std;

void+main()
{
+++++char+ch;
+++++int+x;
+++++cin+>>+ch;
+++++x+=+char+(ar[1]);
+++++cout+<<+x;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|206201|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>#include&lt;iostream&gt;
#include&lt;stdlib&gt;
using namespace std;

void main()
{
 char ch;
 int x;
 cin &gt;&gt; ch;
 x = char (ar[1]);
 cout &lt;&lt; x;
}
</code></pre>

blocks|key|970445|text|只要您只想处理阿拉伯数字，并且在这些数字是连续的编码中工作，并且在与ASCII相同的位置工作，那么所提供的答案就很棒。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|970446|情况几乎总是如此。|970447|如果不是，那么你需要一个合适的库来帮助你。|970448|让我们从ICU开始。|offset|length|970449|970450|首先将字节字符串转换为unicode字符串。(左边作为读者的练习)。然后使用uchar.h查看每个字符。如果|ordered-list-item|970451|为int32_t+u_charDigitValue+++++(+++UChar32++++++c+++++++)|style|CODE|970452|，则值为|970453|unordered-list-item|970454|970455|也许ICU有一些功能可以帮你做--我还没有详细研究过它。|970456|entityMap|0|LINK|mutability|MUTABLE|url|http://site.icu-project.org/|1|http://icu-project.org/apiref/icu4c/uchar_8h.html^0|0|0|0|4|3|0|0|0|12|7|1|0|1|1J|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1A|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1B|8|@]|9|@$H|1C|I|1D|1|1E]]|A|$]]|$1|J|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|1G|8|@]|9|@$H|1H|I|1I|1|1J]]|A|$]]|$1|N|3|O|5|M|7|1K|8|@$H|1L|I|1M|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|6|7|1N|8|@]|9|@]|A|$]]|$1|T|3|-4|5|U|7|1O|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]|16|$5|11|12|13|A|$14|17]]]]

The answers provided are great as long as you only want to handle Arabic numerals, and are working in an encoding where those numerals are sequential, and in the same place as ASCII.

This is almost always the case.

If it isn't then you need a proper library to help you.

Let's start with <a href="http://site.icu-project.org/" rel="nofollow noreferrer">ICU</a>.

<ol>
<li>First convert the byte-string to a unicode string. (Left as an exercise for the reader).</li>
<li>Then use <a href="http://icu-project.org/apiref/icu4c/uchar_8h.html" rel="nofollow noreferrer">uchar.h</a> to look at each character.</li>
<li>if we the character is <code>UBool u_isdigit (UChar32 c)</code></li>
<li>then the value is <code>int32_t u_charDigitValue ( UChar32 c )</code></li>
</ol>

Or maybe ICU has some function to do it for you - I haven't looked at it in detail.

blocks|key|4270819|text|#define+toDigit(c)+(c-'0')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4270820|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>#define toDigit(c) (c-'0')
</code></pre>

blocks|key|4270849|text|您可以使用atoi()函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4270850|#include+<stdlib.h>
#include+<stdio.h>
int+main(int+argc,+char*+argv[]){
++++int+num+;
++++num+=+atoi(argv[1]);
++++printf("\n%25d",+num);
}|code-block|syntax|javascript|4270851|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|$]]

<h3>You can make use of atoi() function</h3>

<pre><code>#include &lt;stdlib.h&gt;
#include &lt;stdio.h&gt;
int main(int argc, char* argv[]){
 int num ;
 num = atoi(argv[1]);
 printf("\n%d", num);
}
</code></pre>

blocks|key|4270876|text|通过这种方式，您可以轻松地将char转换为int，将int转换为char：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4270877|int+charToInt(char+c)
{
+++int+arr[]={0,1,2,3,4,5,6,7,8,9};
+++return+arr[c-'0'];
}|code-block|syntax|javascript|4270878|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|$]]

By this way You can convert char to int and int to char easily:

<pre><code>int charToInt(char c)
{
 int arr[]={0,1,2,3,4,5,6,7,8,9};
 return arr[c-'0'];
}
</code></pre>

blocks|key|206373|text|对我来说，以下方法运行得很好：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|206374|QChar+c+=+'5';
int+x+=+c.digitValue();
//+x+is+now+5|code-block|syntax|javascript|206375|文档：int+QChar::digitValue()+const，它说：|offset|length|206376|206377|返回数字的数值，如果字符不是数字，则返回-1。|blockquote|206378|206379|entityMap|0|LINK|mutability|MUTABLE|url|http://doc.qt.io/qt-5/qchar.html#digitValue^0|0|0|3|T|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@$I|10|J|11|1|12]]|A|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|14|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

For me the following worked quite well:

<pre><code>QChar c = '5';
int x = c.digitValue();
// x is now 5
</code></pre>

Documentation: <a href="http://doc.qt.io/qt-5/qchar.html#digitValue" rel="nofollow noreferrer">int QChar::digitValue() const</a> which says:

<blockquote>
 Returns the numeric value of the digit, or -1 if the character is not
 a digit.
</blockquote>

blocks|key|3423998|text|我同意@jalf。使用sstream库和stoi似乎是推荐的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3423999|#include+<iostream>
#include+<string>
#include+<sstream>

using+namespace+std;

int+main()+{

++stringstream+st;
++st+<<+1++<<+'2';

++cout+<<+stoi(st.str())+%2B+1;+

++return+0;
}|code-block|syntax|javascript|3424000|输出|3424001|13|3424002|我是C%2B%2B的新手，但却是一个长期从事LAMP堆栈开发的人。我希望string类有更多的东西可以在字符和字符串之间平滑地转换，但我还没有找到这样的东西是本机支持的。|3424003|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/sstream/stringstream/|1|https://www.cplusplus.com/reference/string/stoi/^0|B|7|0|K|4|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]]|C|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|16|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

I agree with @jalf. Using the <a href="http://www.cplusplus.com/reference/sstream/stringstream/" rel="nofollow noreferrer">sstream</a> library and <a href="https://www.cplusplus.com/reference/string/stoi/" rel="nofollow noreferrer">stoi</a> appears to be the recommended approach:
<pre class="lang-cpp prettyprint-override"><code>#include &lt;iostream&gt;
#include &lt;string&gt;
#include &lt;sstream&gt;

using namespace std;

int main() {

 stringstream st;
 st &lt;&lt; 1 &lt;&lt; '2';

 cout &lt;&lt; stoi(st.str()) + 1; 

 return 0;
}
</code></pre>
<h1 id="output-98ni">Output</h1>
<pre><code>13
</code></pre>
I am a new student of C++ but a long-time LAMP stack developer. I was hoping the string class had more things to smoothly transition between chars and strings, but I've not yet found such a thing that is natively supported.

I have a string of digits, e.g. "123456789", and I need to extract each one of them to use them in a calculation. I can of course access each char by index, but how do I convert it into an int?

I've looked into atoi(), but it takes a string as argument. Hence I must convert each char into a string and then call atoi on it. Is there a better way?

How to convert a single char into an int

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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我有一串数字，例如"123456789"，我需要提取每一个数字，以便在计算中使用它们。当然，我可以通过索引访问每个字符，但如何将其转换为int？我研究过atoi()，但它接受一个字符串作为参数。因此，我必须将每个字符转换为字符串，然后对其调用atoi。有没有更好的方法？

问如何将单个字符转换为整型
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问如何将单个字符转换为整型EN