blocks|key|1322524|text|对于非变异版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1322525|st+=+myString.substr(0,+myString.size()-1);|code-block|syntax|javascript|1322526|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|$]]

For a non-mutating version:

<pre><code>st = myString.substr(0, myString.size()-1);
</code></pre>

blocks|key|1322886|text|简单的解决方案，如果你使用的是C%2B%2B11。也可能是O(1)时间：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1322887|st.pop_back();|code-block|syntax|javascript|1322888|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|$]]

Simple solution if you are using C++11. Probably O(1) time as well:

<pre><code>st.pop_back();
</code></pre>

blocks|key|1107625|text|if+(str.size+()+>+0)++str.resize+(str.size+()+-+1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1107626|一个std::erase替代是很好的，但我喜欢"-+1“(无论是基于大小还是结束迭代器)-对我来说，它有助于表达意图。|unstyled|1107627|顺便说一句，真的没有std::string::pop_back吗？-看起来很奇怪。|1107628|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>if (str.size () &gt; 0) str.resize (str.size () - 1);
</code></pre>

An std::erase alternative is good, but I like the "- 1" (whether based on a size or end-iterator) - to me, it helps expresses the intent.

BTW - Is there really no std::string::pop_back ? - seems strange.

blocks|key|1107519|text|buf.erase(buf.size()+-+1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1107520|这假设您知道该字符串不为空。如果是这样的话，您将得到一个out_of_range异常。|unstyled|offset|length|style|CODE|1107521|entityMap^0|0|S|C|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>buf.erase(buf.size() - 1);
</code></pre>

This assumes you know that the string is not empty. If so, you'll get an <code>out_of_range</code> exception.

blocks|key|1322929|text|str.erase(+str.end()-1+)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1322930|参考：std::string::erase()+prototype+2|1322931|不需要c%2B%2B11或c%2B%2B0x。|1322932|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/string/string/erase/^0|0|O|0|3|W|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@$9|V|A|W|1|X]]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

<code>str.erase( str.end()-1 )</code>

Reference: <a href="http://www.cplusplus.com/reference/string/string/erase/">std::string::erase() prototype 2</a>

no c++11 or c++0x needed.

blocks|key|1322964|text|这就是你所需要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1322965|#include+<string>++//string::pop_back+&+string::empty

if+(!st.empty())
++++st.pop_back();|code-block|syntax|javascript|1322966|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|$]]

That's all you need:

<pre><code>#include &lt;string&gt; //string::pop_back &amp; string::empty

if (!st.empty())
 st.pop_back();
</code></pre>

blocks|key|1107463|text|int+main+()+{

++string+str1="123";
++string+str2+=+str1.substr+(0,str1.length()-1);

++cout<<str2;+//+output:+12

++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1107464|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>int main () {

 string str1="123";
 string str2 = str1.substr (0,str1.length()-1);

 cout&lt;&lt;str2; // output: 12

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

blocks|key|1322796|text|str.erase(str.begin()+%2B+str.size()+-+1)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1322797|不幸的是，str.erase(str.rbegin())不能编译，因为reverse_iterator不能转换为normal_iterator。|1322798|在这种情况下，C%2B%2B11是您的朋友。|1322799|entityMap^0|0|13|0|5|N|Z|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|K|$]]

<code>str.erase(str.begin() + str.size() - 1)</code>

<code>str.erase(str.rbegin())</code> does not compile unfortunately, since <code>reverse_iterator</code> cannot be converted to a normal_iterator.

C++11 is your friend in this case.

blocks|key|1107745|text|使用C%2B%2B11，您甚至不需要长度/大小。只要字符串不为空，就可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1107746|if+(!st.empty())
++st.erase(std::prev(st.end()));+//+Erase+element+referred+to+by+iterator+one
+++++++++++++++++++++++++++++++++//+before+the+end|code-block|syntax|javascript|1107747|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|$]]

With C++11, you don't even need the length/size. As long as the string is not empty, you can do the following:

<pre><code>if (!st.empty())
 st.erase(std::prev(st.end())); // Erase element referred to by iterator one
 // before the end
</code></pre>

blocks|key|1107911|text|#include<iostream>
using+namespace+std;
int+main(){
++string+s+=+"Hello";//+Here+string+length+is+5+initially
++s[s.length()-1]+=+'\0';+//++marking+the+last+char+to+be+null+character
++s+=+&s[0];+//+using+ampersand+infront+of+the+string+with+index+will+render+a+string+from+the+index+until+null+character+discovered
++cout<<"the+new+length+of+the+string+"<<s+%2B+"+is+"+<<s.length();
++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1107912|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;
using namespace std;
int main(){
 string s = &quot;Hello&quot;;// Here string length is 5 initially
 s[s.length()-1] = '\0'; // marking the last char to be null character
 s = &amp;s[0]; // using ampersand infront of the string with index will render a string from the index until null character discovered
 cout&lt;&lt;&quot;the new length of the string &quot;&lt;&lt;s + &quot; is &quot; &lt;&lt;s.length();
 return 0;
}
</code></pre>

blocks|key|1107903|text|如果长度不为零，您还可以|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1107904|str[str.length()+-+1]+=+'\0';|code-block|syntax|javascript|1107905|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 the length is non zero, you can also

<pre><code>str[str.length() - 1] = '\0';
</code></pre>

How can I remove last character from a C++ string?

I tried <code>st = substr(st.length()-1);</code> But it didn't work.

Remove last character from C++ string

如何从C++字符串中删除最后一个字符？我尝试过st = substr(st.length()-1);，但它不起作用。

问从C++字符串中删除最后一个字符
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