blocks|key|1612076|text|皮奥特S的回答是好的，但是有一种情况它会返回错误的答案，那就是全零的情况：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1612077|000000000000|code-block|syntax|javascript|1612078|若要考虑到这一点，请使用：|1612079|str.erase(0,+min(str.find_first_not_of('0'),+str.size()-1));|1612080|即使str.size()将是0，它也会工作。|style|CODE|1612081|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/25726706/2452869^0|0|7|0|0|0|0|0|2|A|E|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|15|8|@$A|16|B|17|O|P]|$A|18|B|19|O|P]]|9|@]|C|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

<a href="https://stackoverflow.com/a/25726706/2452869">Piotr S's answer</a> is good but there is one case it will return the wrong answer, that is the all-zero case:
<pre><code>000000000000
</code></pre>
To consider this in, use:
<pre><code>str.erase(0, min(str.find_first_not_of('0'), str.size()-1));
</code></pre>
Even when <code>str.size()</code> will be <code>0</code>, it will also work.

blocks|key|644871|text|虽然这可能不是最有效的方法(就运行时速度而言)，但我很想做这样的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|644872|std::cout+<<+std::stoi(strCustData);|code-block|syntax|javascript|644873|当输入完全由0组成时，它提供了一个合理的输出(一个单一的'0')。只有当/如果分析显示这个简单的解决方案是一个问题时，我才会考虑编写更复杂的代码，以期提高速度(我怀疑会出现这种情况)。|offset|length|style|CODE|644874|这里的明显限制是，这确实假定前导零后面的字符是数字。这在您的示例中显然是正确的，但我认为可以想象，您拥有的数据是不正确的。|644875|entityMap^0|0|0|6|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Although it's probably not the most efficient (in terms of run-time speed) way to do things, I'd be tempted to just do something like:

<pre><code>std::cout &lt;&lt; std::stoi(strCustData);
</code></pre>

This is simple and straightforward to use, and gives a reasonable output (a single '0') when the input consists entirely of <code>0</code>s. Only when/if profiling showed that this simple solution was a problem would I consider writing substantially more complex code in the hopes of improving speed (and I doubt that would arise).

The obvious limitation here is that this does assume that the characters after the leading zeros are digits. That's clearly true in your sample, but I suppose it's conceivable that you have data where it's not true.

blocks|key|372245|text|这应该是一个可以应用于任何std::basic_string类型(包括std::string)的泛型函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|372246|template+<typename+T_STR,+typename+T_CHAR>
T_STR+remove_leading(T_STR+const+&+str,+T_CHAR+c)
{
++++auto+end+=+str.end();

++++for+(auto+i+=+str.begin();+i+!=+end;+%2B%2Bi)+{
++++++++if+(*i+!=+c)+{
++++++++++++return+T_STR(i,+end);
++++++++}
++++}

++++//+All+characters+were+leading+or+the+string+is+empty.
++++return+T_STR();
}|code-block|syntax|javascript|372247|在你的例子中，你会像这样使用它：|372248|string+x+=+"0000000057";
string+trimmed+=+remove_leading(x,+'0');

//+trimmed+is+now+"57"|372249|(见演示.)|372250|entityMap|0|LINK|mutability|MUTABLE|url|http://ideone.com/BAejcB^0|D|H|Z|B|0|0|0|0|1|3|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|Q|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

This should work as a generic function that can be applied to any of the <code>std::basic_string</code> types (including <code>std::string</code>):

<pre><code>template &lt;typename T_STR, typename T_CHAR&gt;
T_STR remove_leading(T_STR const &amp; str, T_CHAR c)
{
 auto end = str.end();

 for (auto i = str.begin(); i != end; ++i) {
 if (*i != c) {
 return T_STR(i, end);
 }
 }

 // All characters were leading or the string is empty.
 return T_STR();
}
</code></pre>

In your case you would use it like this:

<pre><code>string x = "0000000057";
string trimmed = remove_leading(x, '0');

// trimmed is now "57"
</code></pre>

(<a href="http://ideone.com/BAejcB" rel="nofollow">See a demo</a>.)

blocks|key|1612107|text|#include<algorithm>
#include<iostream>
#include<string>
using+namespace+std;


int+main()
{
+++string+s="00000001";
+++cout<<s<<endl;
+++int+a=stoi(s);
+++cout<<a;
+++//1+is+ur+ans+zeros+are+removed


}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1612108|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;algorithm&gt;
#include&lt;iostream&gt;
#include&lt;string&gt;
using namespace std;


int main()
{
 string s="00000001";
 cout&lt;&lt;s&lt;&lt;endl;
 int a=stoi(s);
 cout&lt;&lt;a;
 //1 is ur ans zeros are removed


}
</code></pre>

blocks|key|1612113|text|这个C语言友好的代码移除前导零，只要字符数组的长度适合int，就可以使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1612114|char+charArray[6]+=+"000342";
int+inputLen+=+6;

void+stripLeadingZeros()+{
++++int+firstNonZeroDigitAt=0,+inputLen+=+strlen(charArray);

++++//find+location+of+first+non-zero+digit,+if+any
++++while(charArray[firstNonZeroDigitAt]+==+'0')
++++++++firstNonZeroDigitAt%2B%2B;

++++//string+has+no+leading+zeros
++++if(firstNonZeroDigitAt==0)
++++++++return;+

++++//+string+is+all+zeros
++++if(firstNonZeroDigitAt==inputLen)+{
++++++++memset(charArray,+0,+sizeof+charArray);
++++++++strcpy(charArray,+"0");
++++++++inputLen+=+1;
++++++++return;
++++}

++++//shift+the+non-zero+characters+down
++++for(int+j=0;+j<inputLen-firstNonZeroDigitAt;+j%2B%2B)+{
++++++++charArray[j]+=+charArray[firstNonZeroDigitAt%2Bj];
++++}

++++//clear+the+occupied+area+and+update+string+length
++++memset(charArray%2BinputLen-firstNonZeroDigitAt,+0,+inputLen-firstNonZeroDigitAt%2B1);
++++inputLen+-=+firstNonZeroDigitAt;
}|code-block|syntax|javascript|1612115|entityMap^0|R|3|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|$]]

This C language friendly code removes leading zeros and can be used as long as the char array's length fits within an <code>int</code>:

<pre><code>char charArray[6] = "000342";
int inputLen = 6;

void stripLeadingZeros() {
 int firstNonZeroDigitAt=0, inputLen = strlen(charArray);

 //find location of first non-zero digit, if any
 while(charArray[firstNonZeroDigitAt] == '0')
 firstNonZeroDigitAt++;

 //string has no leading zeros
 if(firstNonZeroDigitAt==0)
 return; 

 // string is all zeros
 if(firstNonZeroDigitAt==inputLen) {
 memset(charArray, 0, sizeof charArray);
 strcpy(charArray, "0");
 inputLen = 1;
 return;
 }

 //shift the non-zero characters down
 for(int j=0; j&lt;inputLen-firstNonZeroDigitAt; j++) {
 charArray[j] = charArray[firstNonZeroDigitAt+j];
 }

 //clear the occupied area and update string length
 memset(charArray+inputLen-firstNonZeroDigitAt, 0, inputLen-firstNonZeroDigitAt+1);
 inputLen -= firstNonZeroDigitAt;
}
</code></pre>

I want to remove leading zeroes from string like <code>"0000000057"</code>.

I did like this but didn't get any result:

<pre><code>string AccProcPeriNum = strCustData.substr(pos, 13);

string p48Z03 = AccProcPeriNum.substr(3, 10);
</code></pre>

I want output only 57.

Any idea in C++?

How to remove leading zeros from string using 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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我想从像"0000000057"这样的字符串中移除前导零。我的确喜欢这个，但没有得到任何结果：string AccProcPeriNum = strCustData.substr(pos, 13);string p48Z03 = AccProcPeriNum.substr(3, 10);我只想输出57。在C++有什么想法吗？

问如何使用C++从字符串中删除前导零？
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