我已经阅读了大部分答案，但没有发现任何人使用istringstream

std::string text = "Let me split this into words";

std::istringstream iss(text);
std::vector<std::string> results((std::istream_iterator<std::string>(iss)),
                                 std::istream_iterator<std::string>());

结果是单词的向量，它可以处理有内部空白的字符串，希望这有帮助。

我已经阅读了大部分答案，但没有发现任何人使用istringstream

std::string text = "Let me split this into words";

std::istringstream iss(text);
std::vector<std::string> results((std::istream_iterator<std::string>(iss)),
                                 std::istream_iterator<std::string>());

结果是单词的向量，它可以处理有内部空白的字符串，希望这有帮助。

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));
    }
}

这个好吗?(因为这篇文章完全需要另一个答案:)

string trimBegin(string str)
{
    string whites = "\t\r\n ";
    int i = 0;
    while (whites.find(str[i++]) != whites::npos);
    str.erase(0, i);
    return str;
}

类似的情况下，trimEnd，只是反转极化，指数。

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……

我知道这是一个非常老的问题，但我已经为您的问题添加了几行代码，它从两端删除了空白。

void trim(std::string &line){

    auto val = line.find_last_not_of(" \n\r\t") + 1;

    if(val == line.size() || val == std::string::npos){
        val = line.find_first_not_of(" \n\r\t");
        line = line.substr(val);
    }
    else
        line.erase(val);
}