我目前正在使用以下代码在我的程序中右修剪所有std::字符串:

std::string s;
s.erase(s.find_last_not_of(" \n\r\t")+1);

它工作得很好,但我想知道是否有一些最终情况下它可能会失败?

当然,有优雅的选择和左修剪解决方案的答案是受欢迎的。


当前回答

这里有一个容易理解的解决方案,初学者不习惯编写std::无处不在,还不熟悉常量正确性,迭代器,STL算法等…

#include <string>
#include <cctype> // for isspace
using namespace std;


// Left trim the given string ("  hello!  " --> "hello!  ")
string left_trim(string str) {
    int numStartSpaces = 0;
    for (int i = 0; i < str.length(); i++) {
        if (!isspace(str[i])) break;
        numStartSpaces++;
    }
    return str.substr(numStartSpaces);
}

// Right trim the given string ("  hello!  " --> "  hello!")
string right_trim(string str) {
    int numEndSpaces = 0;
    for (int i = str.length() - 1; i >= 0; i--) {
        if (!isspace(str[i])) break;
        numEndSpaces++;
    }
    return str.substr(0, str.length() - numEndSpaces);
}

// Left and right trim the given string ("  hello!  " --> "hello!")
string trim(string str) {
    return right_trim(left_trim(str));
}

希望能有所帮助……

其他回答

这太烦人了,我

必须谷歌它 发现我必须使用火箭科学 字符串中没有简单的trim/toupper函数

对我来说,这是最快的解决方法:

CString tmp(line.c_str());
tmp = tmp.Trim().MakeLower();
string buffer = tmp;

我可以使用lambda ops,迭代器,以及所有的东西,这很酷。但我只需要处理一个字符串而不是一个字符…

s.erase(0, s.find_first_not_of(" \n\r\t"));                                                                                               
s.erase(s.find_last_not_of(" \n\r\t")+1);   

我认为在这个例子中使用宏是一个很好的实践:(适用于c++ 98)

#define TRIM_CHARACTERS " \t\n\r\f\v"
#define TRIM_STRING(given) \
    given.erase(given.find_last_not_of(TRIM_CHARACTERS) + 1); \
    given.erase(0, given.find_first_not_of(TRIM_CHARACTERS));

例子:

#include <iostream>
#include <string>

#define TRIM_CHARACTERS " \t\n\r\f\v"
#define TRIM_STRING(given) \
    given.erase(given.find_last_not_of(TRIM_CHARACTERS) + 1); \
    given.erase(0, given.find_first_not_of(TRIM_CHARACTERS));

int main(void) {
  std::string text("  hello world!! \t  \r");
  TRIM_STRING(text);
  std::cout << text; // "hello world!!"
}

For what it's worth, here is a trim implementation with an eye towards performance. It's much quicker than many other trim routines I've seen around. Instead of using iterators and std::finds, it uses raw c strings and indices. It optimizes the following special cases: size 0 string (do nothing), string with no whitespace to trim (do nothing), string with only trailing whitespace to trim (just resize the string), string that's entirely whitespace (just clear the string). And finally, in the worst case (string with leading whitespace), it does its best to perform an efficient copy construction, performing only 1 copy and then moving that copy in place of the original string.

void TrimString(std::string & str)
{ 
    if(str.empty())
        return;

    const auto pStr = str.c_str();

    size_t front = 0;
    while(front < str.length() && std::isspace(int(pStr[front]))) {++front;}

    size_t back = str.length();
    while(back > front && std::isspace(int(pStr[back-1]))) {--back;}

    if(0 == front)
    {
        if(back < str.length())
        {
            str.resize(back - front);
        }
    }
    else if(back <= front)
    {
        str.clear();
    }
    else
    {
        str = std::move(std::string(str.begin()+front, str.begin()+back));
    }
}

C++11:

int i{};
string s = " h e ll \t\n  o";
string trim = " \n\t";

while ((i = s.find_first_of(trim)) != -1)
    s.erase(i,1);

cout << s;

输出:

hello

也适用于空字符串