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);
}
根据Galik的回答,我做了这个。这大部分都在这里,所以我不必一遍又一遍地写。C++仍然没有原生拆分函数,这真是太疯狂了。特征:
应该很快。容易理解(我认为)。合并空节。使用多个分隔符(例如“\r\n”)很简单
#include <string>
#include <vector>
#include <algorithm>
std::vector<std::string> split(const std::string& s, const std::string& delims)
{
using namespace std;
vector<string> v;
// Start of an element.
size_t elemStart = 0;
// We start searching from the end of the previous element, which
// initially is the start of the string.
size_t elemEnd = 0;
// Find the first non-delim, i.e. the start of an element, after the end of the previous element.
while((elemStart = s.find_first_not_of(delims, elemEnd)) != string::npos)
{
// Find the first delem, i.e. the end of the element (or if this fails it is the end of the string).
elemEnd = s.find_first_of(delims, elemStart);
// Add it.
v.emplace_back(s, elemStart, elemEnd == string::npos ? string::npos : elemEnd - elemStart);
}
// When there are no more non-spaces, we are done.
return v;
}
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。
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!