如何使用 C++ 从字符串中删除前导零？
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How to remove leading zeros from string using C++?
提问by samio peter
I want to remove leading zeroes from string like "0000000057".
我想从字符串中删除前导零，如"0000000057".
I did like this but didn't get any result:
我确实喜欢这个，但没有得到任何结果：
string AccProcPeriNum = strCustData.substr(pos, 13);
string p48Z03 = AccProcPeriNum.substr(3, 10);
I want output only 57.
我只想要输出57。
Any idea in C++?
在 C++ 中有什么想法吗？
回答by intijk
Piotr S's answer is good but there is one case it will return wrong answer, that is the all zero case:
Piotr S 的回答很好，但有一种情况会返回错误答案，即全零情况：
000000000000
000000000000
To consider this in, use:
要考虑这一点，请使用：
str.erase(0, min(str.find_first_not_of('0'), str.size()-1));
Even you str.size() be 0, it will also work.
即使你的 str.size() 为 0，它也能工作。
回答by Piotr Skotnicki
回答by Jerry Coffin
Although it's probably notthe most efficient (in terms of run-time speed) way to do things, I'd be tempted to just do something like:
尽管这可能不是最有效的（就运行时速度而言）做事的方式，但我很想这样做：
std::cout << std::stoi(strCustData);
This is simple and straightforward to use, and gives a reasonable output (a single '0') when the input consists entirely of 0s. 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).
这使用起来简单直接，并且在输入完全由0s组成时给出了合理的输出（单个“0”）。只有当/如果分析表明这个简单的解决方案是一个问题，我才会考虑编写更复杂的代码以希望提高速度（我怀疑会出现这种情况）。
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.
这里明显的限制是，这确实假设前导零后面的字符是数字。这在您的样本中显然是正确的，但我想可以想象您拥有不正确的数据。
回答by cdhowie
This should work as a generic function that can be applied to any of the std::basic_stringtypes (including std::string):
这应该作为可以应用于任何std::basic_string类型（包括std::string）的通用函数：
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; ++i) {
if (*i != c) {
return T_STR(i, end);
}
}
// All characters were leading or the string is empty.
return T_STR();
}
In your case you would use it like this:
在你的情况下，你会像这样使用它：
string x = "0000000057";
string trimmed = remove_leading(x, '0');
// trimmed is now "57"
(See a demo.)
（见演示。）
回答by Mohit Jawale
#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
}
回答by Yoga
This C language friendly code removes leading zeros and can be used as long as the char array's length fits within an int:
这个 C 语言友好的代码删除了前导零，只要字符数组的长度在一个范围内就可以使用int：
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<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;
}
