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);
}
如果您希望按某些字符分割字符串,可以使用
#include<iostream>
#include<string>
#include<vector>
#include<iterator>
#include<sstream>
#include<string>
using namespace std;
void replaceOtherChars(string &input, vector<char> ÷rs)
{
const char divider = dividers.at(0);
int replaceIndex = 0;
vector<char>::iterator it_begin = dividers.begin()+1,
it_end= dividers.end();
for(;it_begin!=it_end;++it_begin)
{
replaceIndex = 0;
while(true)
{
replaceIndex=input.find_first_of(*it_begin,replaceIndex);
if(replaceIndex==-1)
break;
input.at(replaceIndex)=divider;
}
}
}
vector<string> split(string str, vector<char> chars, bool missEmptySpace =true )
{
vector<string> result;
const char divider = chars.at(0);
replaceOtherChars(str,chars);
stringstream stream;
stream<<str;
string temp;
while(getline(stream,temp,divider))
{
if(missEmptySpace && temp.empty())
continue;
result.push_back(temp);
}
return result;
}
int main()
{
string str ="milk, pigs.... hot-dogs ";
vector<char> arr;
arr.push_back(' '); arr.push_back(','); arr.push_back('.');
vector<string> result = split(str,arr);
vector<string>::iterator it_begin= result.begin(),
it_end= result.end();
for(;it_begin!=it_end;++it_begin)
{
cout<<*it_begin<<endl;
}
return 0;
}
这是一个顶级答案的扩展。它现在支持设置返回元素的最大数量N。字符串的最后一位将在第N个元素中结束。MAXELEMENTS参数是可选的,如果设置为默认值0,它将返回无限数量的元素。:-)
.h:
class Myneatclass {
public:
static std::vector<std::string>& split(const std::string &s, char delim, std::vector<std::string> &elems, const size_t MAXELEMENTS = 0);
static std::vector<std::string> split(const std::string &s, char delim, const size_t MAXELEMENTS = 0);
};
.cpp:
std::vector<std::string>& Myneatclass::split(const std::string &s, char delim, std::vector<std::string> &elems, const size_t MAXELEMENTS) {
std::stringstream ss(s);
std::string item;
while (std::getline(ss, item, delim)) {
elems.push_back(item);
if (MAXELEMENTS > 0 && !ss.eof() && elems.size() + 1 >= MAXELEMENTS) {
std::getline(ss, item);
elems.push_back(item);
break;
}
}
return elems;
}
std::vector<std::string> Myneatclass::split(const std::string &s, char delim, const size_t MAXELEMENTS) {
std::vector<std::string> elems;
split(s, delim, elems, MAXELEMENTS);
return elems;
}
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。
最小的解决方案是一个函数,它将std::字符串和一组分隔符(作为std::string)作为输入,并返回std:::字符串的std::向量。
#include <string>
#include <vector>
std::vector<std::string>
tokenize(const std::string& str, const std::string& delimiters)
{
using ssize_t = std::string::size_type;
const ssize_t str_ln = str.length();
ssize_t last_pos = 0;
// container for the extracted tokens
std::vector<std::string> tokens;
while (last_pos < str_ln) {
// find the position of the next delimiter
ssize_t pos = str.find_first_of(delimiters, last_pos);
// if no delimiters found, set the position to the length of string
if (pos == std::string::npos)
pos = str_ln;
// if the substring is nonempty, store it in the container
if (pos != last_pos)
tokens.emplace_back(str.substr(last_pos, pos - last_pos));
// scan past the previous substring
last_pos = pos + 1;
}
return tokens;
}
用法示例:
#include <iostream>
int main()
{
std::string input_str = "one + two * (three - four)!!---! ";
const char* delimiters = "! +- (*)";
std::vector<std::string> tokens = tokenize(input_str, delimiters);
std::cout << "input = '" << input_str << "'\n"
<< "delimiters = '" << delimiters << "'\n"
<< "nr of tokens found = " << tokens.size() << std::endl;
for (const std::string& tk : tokens) {
std::cout << "token = '" << tk << "'\n";
}
return 0;
}
我用这个分隔符分隔字符串。第一个将结果放入预先构建的向量中,第二个返回新向量。
#include <string>
#include <sstream>
#include <vector>
#include <iterator>
template <typename Out>
void split(const std::string &s, char delim, Out result) {
std::istringstream iss(s);
std::string item;
while (std::getline(iss, item, delim)) {
*result++ = item;
}
}
std::vector<std::string> split(const std::string &s, char delim) {
std::vector<std::string> elems;
split(s, delim, std::back_inserter(elems));
return elems;
}
请注意,此解决方案不会跳过空令牌,因此下面将找到4项,其中一项为空:
std::vector<std::string> x = split("one:two::three", ':');