blocks|key|287874|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|287875|在C%2B%2B11中，使用std::stoi作为：|unordered-list-item|offset|length|style|CODE|287876|287877|std::string+s=+"10"；int+i=+std::stoi(s)；|287878|请注意，如果无法执行转换，std::stoi将抛出std::invalid_argument类型的异常，或者如果转换导致溢出(即字符串值对于int类型太大)，则抛出std::out_of_range异常。您可以使用std::stol或std:stoll，以防int对于输入字符串而言太小。|287879|287880|在C%2B%2B03/98中，可以使用以下任何选项：|287881|287882|std::string+s=+"+10+"；int+i；//接近一个std：：istringstream+>>+i；//这之后//接近两个sscanf(s.c_str()，"%25d"，&i)；//这个之后，i是10|287883|请注意，对于输入s+=+"10jh"，上述两种方法将失败。它们将返回10，而不是通知错误。因此，安全和健壮的方法是编写自己的函数来解析输入字符串，并验证每个字符以检查它是否是数字，然后进行相应的工作。这里有一个健壮的实现(虽然没有经过测试)：|287884|int+to_int(char+const+*s)
{
+++++if+(+s+==+NULL+%7C%7C+*s+==+'\0'+)
++++++++throw+std::invalid_argument("null+or+empty+string+argument");

+++++bool+negate+=+(s[0]+==+'-');
 + + if+(+*s+==+'%2B'+%7C%7C+*s+==+'-'+) 
 + + + ++%2B%2Bs;

+++++if+(+*s+==+'\0')
++++++++throw+std::invalid_argument("sign+character+only.");

 + + int+result+=+0;
 + + while(*s)
 + + {
 + + + + +if+(+*s+<+'0'+%7C%7C+*s+>+'9'+)
++++++++++++throw+std::invalid_argument("invalid+input+string");
 + + + + +result+=+result+*+10+ -+(*s+-+'0');+ //assume+negative+number
 + + + + +%2B%2Bs;
 + + }
 + + return+negate+?+result+:+-result;+//-result+is+positive!
} |code-block|syntax|javascript|287885|此解决方案是my+another+solution的略微修改版本。|287886|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/string/basic_string/stol|1|2|3|https://stackoverflow.com/questions/9620437/string-const-char-size-t-to-int^0|0|A|9|A|9|0|0|0|0|D|9|P|L|1Y|3|2A|H|2Z|9|39|9|3L|3|2Z|9|1|39|9|2|0|0|0|0|0|8|A|0|0|6|J|3|0^^$0|@$1|2|3|-4|4|5|6|1D|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1E|7|@$D|1F|E|1G|F|G]]|8|@$D|1H|E|1I|1|1J]]|9|$]]|$1|H|3|-4|4|5|6|1K|7|@]|8|@]|9|$]]|$1|I|3|J|4|5|6|1L|7|@]|8|@]|9|$]]|$1|K|3|L|4|5|6|1M|7|@$D|1N|E|1O|F|G]|$D|1P|E|1Q|F|G]|$D|1R|E|1S|F|G]|$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]|$D|1Z|E|20|F|G]]|8|@$D|21|E|22|1|23]|$D|24|E|25|1|26]]|9|$]]|$1|M|3|-4|4|5|6|27|7|@]|8|@]|9|$]]|$1|N|3|O|4|C|6|28|7|@]|8|@]|9|$]]|$1|P|3|-4|4|5|6|29|7|@]|8|@]|9|$]]|$1|Q|3|R|4|5|6|2A|7|@]|8|@]|9|$]]|$1|S|3|T|4|5|6|2B|7|@$D|2C|E|2D|F|G]]|8|@]|9|$]]|$1|U|3|V|4|W|6|2E|7|@]|8|@]|9|$X|Y]]|$1|Z|3|10|4|5|6|2F|7|@]|8|@$D|2G|E|2H|1|2I]]|9|$]]|$1|11|3|-4|4|5|6|2J|7|@]|8|@]|9|$]]]|12|$13|$4|14|15|16|9|$17|18]]|19|$4|14|15|16|9|$17|18]]|1A|$4|14|15|16|9|$17|18]]|1B|$4|14|15|16|9|$17|1C]]]]

<ul>
<li>In C++11, use <a href="http://en.cppreference.com/w/cpp/string/basic_string/stol" rel="noreferrer"><code>std::stoi</code></a> as:
<pre><code> std::string s = &quot;10&quot;;
 int i = std::stoi(s);
</code></pre>
Note that <code>std::stoi</code> will throw exception of type <code>std::invalid_argument</code> if the conversion cannot be performed, or <code>std::out_of_range</code> if the conversion results in overflow(i.e when the string value is too big for <code>int</code> type). You can use <a href="http://en.cppreference.com/w/cpp/string/basic_string/stol" rel="noreferrer"><code>std::stol</code></a> or <a href="http://en.cppreference.com/w/cpp/string/basic_string/stol" rel="noreferrer"><code>std:stoll</code></a> though in case <code>int</code> seems too small for the input string.
</li>
<li>In C++03/98, any of the following can be used:
<pre><code> std::string s = &quot;10&quot;;
 int i;

 //approach one
 std::istringstream(s) &gt;&gt; i; //i is 10 after this

 //approach two
 sscanf(s.c_str(), &quot;%d&quot;, &amp;i); //i is 10 after this
</code></pre>
</li>
</ul>
Note that the above two approaches would fail for input <code>s = &quot;10jh&quot;</code>. They will return 10 instead of notifying error. So the safe and robust approach is to write your own function that parses the input string, and verify each character to check if it is digit or not, and then work accordingly. Here is one robust implemtation (untested though):
<pre><code>int to_int(char const *s)
{
 if ( s == NULL || *s == '\0' )
 throw std::invalid_argument(&quot;null or empty string argument&quot;);

 bool negate = (s[0] == '-');
     if ( *s == '+' || *s == '-' ) 
        ++s;

 if ( *s == '\0')
 throw std::invalid_argument(&quot;sign character only.&quot;);

     int result = 0;
     while(*s)
     {
          if ( *s &lt; '0' || *s &gt; '9' )
 throw std::invalid_argument(&quot;invalid input string&quot;);
          result = result * 10  - (*s - '0');  //assume negative number
          ++s;
     }
     return negate ? result : -result; //-result is positive!
} 
</code></pre>
This solution is slightly modified version of <a href="https://stackoverflow.com/questions/9620437/string-const-char-size-t-to-int">my another solution</a>.

blocks|key|287888|text|您可以使用istringstream。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|287889|string+s+=+"10";

//+create+an+input+stream+with+your+string.
istringstream+is(str);

int+i;
//+use+is+like+an+input+stream
is+>>+i;|code-block|syntax|javascript|287890|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/iostream/istringstream/^0|5|D|5|D|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You can use <a href="http://www.cplusplus.com/reference/iostream/istringstream/" rel="nofollow noreferrer"><code>istringstream</code></a>.
<pre><code>string s = &quot;10&quot;;

// create an input stream with your string.
istringstream is(str);

int i;
// use is like an input stream
is &gt;&gt; i;
</code></pre>

blocks|key|288735|text|您可以使用boost::lexical_cast|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|288736|#include+<iostream>
#include+<boost/lexical_cast.hpp>

int+main(+int+argc,+char*+argv[]+){
std::string+s1+=+"10";
std::string+s2+=+"abc";
int+i;

+++try+++{
++++++i+=+boost::lexical_cast<int>(+s1+);
+++}
+++catch(+boost::bad_lexical_cast+&+e+){
++++++std::cout+<<+"Exception+caught+:+"+<<+e.what()+<<+std::endl;
+++}

+++try+++{
++++++i+=+boost::lexical_cast<int>(+s2+);
+++}
+++catch(+boost::bad_lexical_cast+&+e+){
++++++std::cout+<<+"Exception+caught+:+"+<<+e.what()+<<+std::endl;
+++}

+++return+0;
}|code-block|syntax|javascript|288737|entityMap|0|LINK|mutability|MUTABLE|url|http://www.boost.org/doc/libs/1_45_0/libs/conversion/lexical_cast.htm^0|5|J|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can use <a href="http://www.boost.org/doc/libs/1_45_0/libs/conversion/lexical_cast.htm" rel="noreferrer">boost::lexical_cast</a>:

<pre><code>#include &lt;iostream&gt;
#include &lt;boost/lexical_cast.hpp&gt;

int main( int argc, char* argv[] ){
std::string s1 = "10";
std::string s2 = "abc";
int i;

 try {
 i = boost::lexical_cast&lt;int&gt;( s1 );
 }
 catch( boost::bad_lexical_cast &amp; e ){
 std::cout &lt;&lt; "Exception caught : " &lt;&lt; e.what() &lt;&lt; std::endl;
 }

 try {
 i = boost::lexical_cast&lt;int&gt;( s2 );
 }
 catch( boost::bad_lexical_cast &amp; e ){
 std::cout &lt;&lt; "Exception caught : " &lt;&lt; e.what() &lt;&lt; std::endl;
 }

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

blocks|key|3004848|text|没有“正确的方法”。如果您想要一个通用的(但不是最优的)解决方案，您可以使用boost::lexical+cast。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3004849|C%2B%2B的一种常见解决方案是使用std::ostream和<<+operator。您可以使用stringstream和stringstream::str()方法来转换为字符串。|3004850|如果你真的需要一种快速的机制(记住20/80规则)，你可以寻找一个“专用”的解决方案，比如C%2B%2B+String+Toolkit+Library|3004851|诚挚的问候,|3004852|Marcin|3004853|entityMap|0|LINK|mutability|MUTABLE|url|http://www.partow.net/programming/strtk/index.html^0|12|J|0|F|C|S|B|19|C|1M|J|0|19|Q|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|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|H|3|I|5|6|7|17|8|@]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|J|3|K|5|6|7|1B|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1C|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

There is no &quot;right way&quot;. If you want a universal (but suboptimal) solution you can use a <code>boost::lexical cast</code>.
A common solution for C++ is to use <code>std::ostream</code> and <code>&lt;&lt; operator</code>. You can use a <code>stringstream</code> and <code>stringstream::str()</code> method for conversion to string.
If you really require a fast mechanism (remember the 20/80 rule) you can look for a &quot;dedicated&quot; solution like <a href="http://www.partow.net/programming/strtk/index.html" rel="nofollow noreferrer">C++ String Toolkit Library</a>
Best Regards, 
Marcin

blocks|key|3218522|text|一些方便快捷的函数(如果您没有使用Boost)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3218523|template<typename+T>
std::string+ToString(const+T&+v)
{
++++std::ostringstream+ss;
++++ss+<<+v;
++++return+ss.str();
}

template<typename+T>
T+FromString(const+std::string&+str)
{
++++std::istringstream+ss(str);
++++T+ret;
++++ss+>>+ret;
++++return+ret;
}|code-block|syntax|javascript|3218524|示例：|3218525|int+i+=+FromString<int>(s);
std::string+str+=+ToString(i);|3218526|适用于任何可流式处理的类型(浮点数等)。您将需要#include+<sstream>，可能还需要#include+<string>。|offset|length|style|CODE|3218527|entityMap^0|0|0|0|0|O|I|1C|H|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Some handy quick functions (if you're not using Boost):

<pre><code>template&lt;typename T&gt;
std::string ToString(const T&amp; v)
{
 std::ostringstream ss;
 ss &lt;&lt; v;
 return ss.str();
}

template&lt;typename T&gt;
T FromString(const std::string&amp; str)
{
 std::istringstream ss(str);
 T ret;
 ss &gt;&gt; ret;
 return ret;
}
</code></pre>

Example:

<pre><code>int i = FromString&lt;int&gt;(s);
std::string str = ToString(i);
</code></pre>

Works for any streamable types (floats etc). You'll need to <code>#include &lt;sstream&gt;</code> and possibly also <code>#include &lt;string&gt;</code>.

<blockquote>
 Possible Duplicate: 
 <a href="https://stackoverflow.com/questions/194465/how-to-parse-a-string-to-an-int-in-c">How to parse a string to an int in C++?</a> 
</blockquote>



I have done some research and some people say to use atio and others say it's bad, and I can't get it to work anyways.

So I just want to ask flat out, whats the right way to convert a string to a int.

<pre><code>string s = "10";
int i = s....?
</code></pre>

Thanks!

c++ parse int from string

 可能重复：我做了一些研究，一些人说使用atio，另一些人说它很糟糕，我无论如何都不能让它工作。所以我只想直截了当地问，把字符串转换成整型的正确方法是什么。string s = "10";int i = s....?谢谢!

问c++从字符串解析整型
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