穷人的绳子装饰(仅限空格):

std::string trimSpaces(const std::string& str)
{
    int start, len;
    
    for (start = 0; start < str.size() && str[start] == ' '; start++);
    for (len = str.size() - start; len > 0 && str[start + len - 1] == ' '; len--);
    
    return str.substr(start, len);
}

为什么不用?

auto no_space = [](char ch) -> bool {
  return !std::isspace<char>(ch, std::locale::classic());
};
auto ltrim = [](std::string& s) -> std::string& {
  s.erase(s.begin(), std::find_if(s.begin(), s.end(), no_space));
  return s;
};
auto rtrim = [](std::string& s) -> std::string& {
  s.erase(std::find_if(s.rbegin(), s.rend(), no_space).base(), s.end());
  return s;
};
auto trim_copy = [](std::string s) -> std::string& { return ltrim(rtrim(s)); };
auto trim = [](std::string& s) -> std::string& { return ltrim(rtrim(s)); };

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;
}

从Cplusplus.com上窃取的

std::string choppa(const std::string &t, const std::string &ws)
{
    std::string str = t;
    size_t found;
    found = str.find_last_not_of(ws);
    if (found != std::string::npos)
        str.erase(found+1);
    else
        str.clear();            // str is all whitespace

    return str;
}

这也适用于空情况。: -)

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

以下是我的看法:

size_t beg = s.find_first_not_of(" \r\n");
return (beg == string::npos) ? "" : in.substr(beg, s.find_last_not_of(" \r\n") - beg);