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!!"
}

使用Boost的字符串算法是最简单的:

#include <boost/algorithm/string.hpp>

std::string str("hello world! ");
boost::trim_right(str);

STR现在是"hello world!"。还有trim_left和trim，它们修剪两边。

如果你给上面的函数名加上_copy后缀，例如trim_copy，函数将返回一个经过修剪的字符串副本，而不是通过引用修改它。

如果你给上面的任何函数名加上_if后缀，例如trim_copy_if，你可以修剪所有满足自定义谓词的字符，而不是只有空白。

我的解决方案是基于@比尔蜥蜴的回答。

注意，如果输入字符串只包含空格，这些函数将返回空字符串。

const std::string StringUtils::WHITESPACE = " \n\r\t";

std::string StringUtils::Trim(const std::string& s)
{
    return TrimRight(TrimLeft(s));
}

std::string StringUtils::TrimLeft(const std::string& s)
{
    size_t startpos = s.find_first_not_of(StringUtils::WHITESPACE);
    return (startpos == std::string::npos) ? "" : s.substr(startpos);
}

std::string StringUtils::TrimRight(const std::string& s)
{
    size_t endpos = s.find_last_not_of(StringUtils::WHITESPACE);
    return (endpos == std::string::npos) ? "" : s.substr(0, endpos+1);
}

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……

我喜欢tzaman的解决方案，唯一的问题是它不修剪只包含空格的字符串。

为了纠正这1个缺陷，在2个微调行之间添加str.clear()

std::stringstream trimmer;
trimmer << str;
str.clear();
trimmer >> str;

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));
    }
}