如何迭代由空格分隔的单词组成的字符串中的单词?
注意,我对C字符串函数或那种字符操作/访问不感兴趣。比起效率,我更喜欢优雅。我当前的解决方案:
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
int main() {
string s = "Somewhere down the road";
istringstream iss(s);
do {
string subs;
iss >> subs;
cout << "Substring: " << subs << endl;
} while (iss);
}
作为一个业余爱好者,这是我想到的第一个解决方案。我有点好奇,为什么我还没有在这里看到类似的解决方案,是不是我的做法有根本问题?
#include <iostream>
#include <string>
#include <vector>
std::vector<std::string> split(const std::string &s, const std::string &delims)
{
std::vector<std::string> result;
std::string::size_type pos = 0;
while (std::string::npos != (pos = s.find_first_not_of(delims, pos))) {
auto pos2 = s.find_first_of(delims, pos);
result.emplace_back(s.substr(pos, std::string::npos == pos2 ? pos2 : pos2 - pos));
pos = pos2;
}
return result;
}
int main()
{
std::string text{"And then I said: \"I don't get it, why would you even do that!?\""};
std::string delims{" :;\".,?!"};
auto words = split(text, delims);
std::cout << "\nSentence:\n " << text << "\n\nWords:";
for (const auto &w : words) {
std::cout << "\n " << w;
}
return 0;
}
http://cpp.sh/7wmzy
我已经使用strtok滚动了自己的代码,并使用boost拆分了一个字符串。我找到的最好的方法是C++字符串工具包库。它非常灵活和快速。
#include <iostream>
#include <vector>
#include <string>
#include <strtk.hpp>
const char *whitespace = " \t\r\n\f";
const char *whitespace_and_punctuation = " \t\r\n\f;,=";
int main()
{
{ // normal parsing of a string into a vector of strings
std::string s("Somewhere down the road");
std::vector<std::string> result;
if( strtk::parse( s, whitespace, result ) )
{
for(size_t i = 0; i < result.size(); ++i )
std::cout << result[i] << std::endl;
}
}
{ // parsing a string into a vector of floats with other separators
// besides spaces
std::string s("3.0, 3.14; 4.0");
std::vector<float> values;
if( strtk::parse( s, whitespace_and_punctuation, values ) )
{
for(size_t i = 0; i < values.size(); ++i )
std::cout << values[i] << std::endl;
}
}
{ // parsing a string into specific variables
std::string s("angle = 45; radius = 9.9");
std::string w1, w2;
float v1, v2;
if( strtk::parse( s, whitespace_and_punctuation, w1, v1, w2, v2) )
{
std::cout << "word " << w1 << ", value " << v1 << std::endl;
std::cout << "word " << w2 << ", value " << v2 << std::endl;
}
}
return 0;
}
该工具包比这个简单示例显示的灵活性要高得多,但它在将字符串解析为有用元素方面的实用性令人难以置信。
如果您需要通过非空格符号解析字符串,则字符串流可能很方便:
string s = "Name:JAck; Spouse:Susan; ...";
string dummy, name, spouse;
istringstream iss(s);
getline(iss, dummy, ':');
getline(iss, name, ';');
getline(iss, dummy, ':');
getline(iss, spouse, ';')
对于一个大得离谱而且可能是冗余的版本,可以尝试很多For循环。
string stringlist[10];
int count = 0;
for (int i = 0; i < sequence.length(); i++)
{
if (sequence[i] == ' ')
{
stringlist[count] = sequence.substr(0, i);
sequence.erase(0, i+1);
i = 0;
count++;
}
else if (i == sequence.length()-1) // Last word
{
stringlist[count] = sequence.substr(0, i+1);
}
}
它并不漂亮,但总的来说(除了标点符号和一系列其他错误)它是有效的!
这是我的版本获取了Kev的来源:
#include <string>
#include <vector>
void split(vector<string> &result, string str, char delim ) {
string tmp;
string::iterator i;
result.clear();
for(i = str.begin(); i <= str.end(); ++i) {
if((const char)*i != delim && i != str.end()) {
tmp += *i;
} else {
result.push_back(tmp);
tmp = "";
}
}
}
之后,调用函数并执行以下操作:
vector<string> hosts;
split(hosts, "192.168.1.2,192.168.1.3", ',');
for( size_t i = 0; i < hosts.size(); i++){
cout << "Connecting host : " << hosts.at(i) << "..." << endl;
}