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));
    }
}

下面是一个用正则表达式进行修整的解决方案

#include <string>
#include <regex>

string trim(string str){
    return regex_replace(str, regex("(^[ ]+)|([ ]+$)"),"");
}

试试这个，对我很管用。

inline std::string trim(std::string& str)
{
    str.erase(str.find_last_not_of(' ')+1);         //suffixing spaces
    str.erase(0, str.find_first_not_of(' '));       //prefixing spaces
    return str;
}

在空字符串的情况下，你的代码假设将1添加到string::npos得到0。String::npos的类型是String::size_type，无符号。因此，您依赖于加法的溢出行为。

从c++17开始，标准库的一些部分被删除了。幸运的是，从c++11开始，我们有了lambdas，这是一个更好的解决方案。

#include <algorithm> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(), [](unsigned char ch) {
        return !std::isspace(ch);
    }));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(), [](unsigned char ch) {
        return !std::isspace(ch);
    }).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    rtrim(s);
    ltrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

感谢https://stackoverflow.com/a/44973498/524503提供的现代解决方案。

最初的回答:

我倾向于使用这3种中的一种来满足我的装饰需求:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start
static inline std::string &ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
    return s;
}

// trim from end
static inline std::string &rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
    return s;
}

// trim from both ends
static inline std::string &trim(std::string &s) {
    return ltrim(rtrim(s));
}

它们是相当不言自明的，而且工作得非常好。

编辑:顺便说一句，我有std::ptr_fun在那里，以帮助消除std::isspace的歧义，因为实际上有第二个定义支持区域设置。这本来也可以是一个石膏，但我更喜欢这个。

编辑:处理一些关于通过引用接受参数、修改和返回参数的注释。我同意。我可能更喜欢的实现是两组函数，一组用于到位，另一组用于复制。一个更好的例子是:

#include <algorithm> 
#include <functional> 
#include <cctype>
#include <locale>

// trim from start (in place)
static inline void ltrim(std::string &s) {
    s.erase(s.begin(), std::find_if(s.begin(), s.end(),
            std::not1(std::ptr_fun<int, int>(std::isspace))));
}

// trim from end (in place)
static inline void rtrim(std::string &s) {
    s.erase(std::find_if(s.rbegin(), s.rend(),
            std::not1(std::ptr_fun<int, int>(std::isspace))).base(), s.end());
}

// trim from both ends (in place)
static inline void trim(std::string &s) {
    rtrim(s);
    ltrim(s);
}

// trim from start (copying)
static inline std::string ltrim_copy(std::string s) {
    ltrim(s);
    return s;
}

// trim from end (copying)
static inline std::string rtrim_copy(std::string s) {
    rtrim(s);
    return s;
}

// trim from both ends (copying)
static inline std::string trim_copy(std::string s) {
    trim(s);
    return s;
}

我保留了上面的原始答案，但是为了上下文和保持高投票的答案仍然可用。

这个……怎么样?

#include <iostream>
#include <string>
#include <regex>

std::string ltrim( std::string str ) {
    return std::regex_replace( str, std::regex("^\\s+"), std::string("") );
}

std::string rtrim( std::string str ) {
    return std::regex_replace( str, std::regex("\\s+$"), std::string("") );
}

std::string trim( std::string str ) {
    return ltrim( rtrim( str ) );
}

int main() {

    std::string str = "   \t  this is a test string  \n   ";
    std::cout << "-" << trim( str ) << "-\n";
    return 0;

}

注意:我对c++还是个新手，所以如果我在这里离题了，请原谅。