blocks|key|1437041|text|或者，您可以使用“正确的”方法，类似于原始的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|1437042|(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)|1437043|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|$]]

<p>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)</p>

<p>(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)</p>


blocks|key|1437170|text|下面的方法有什么问题吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1437171|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|1437172|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|$]]

<p>Any problems with the following way of doing it?</p>

<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|1437190|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|1437191|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|1437209|text|#define+toDigit(c)+(c-'0')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1437210|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|1437252|text|您可以使用atoi()函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1437253|#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|1437254|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|1651825|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|1651826|int+charToInt(char+c)
{
+++int+arr[]={0,1,2,3,4,5,6,7,8,9};
+++return+arr[c-'0'];
}|code-block|syntax|javascript|1651827|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|$]]

<p>By this way You can convert char to int and int to char easily:</p>

<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|1437348|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|1437349|#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|1437350|输出|1437351|13|1437352|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。|1437353|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]]]]

<p>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:</p>
<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>
<p>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.</p>


<p>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?</p>

<p>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?</p>


How to convert a single char into an int

LAMP堆栈

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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