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);
}
使用std::string_view和Eric Niebler的range-v3库:
https://wandbox.org/permlink/kW5lwRCL1pxjp2pW
#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"
#include "range/v3/algorithm.hpp"
int main() {
std::string s = "Somewhere down the range v3 library";
ranges::for_each(s
| ranges::view::split(' ')
| ranges::view::transform([](auto &&sub) {
return std::string_view(&*sub.begin(), ranges::distance(sub));
}),
[](auto s) {std::cout << "Substring: " << s << "\n";}
);
}
通过使用循环的范围而不是范围::for_each算法:
#include <iostream>
#include <string>
#include <string_view>
#include "range/v3/view.hpp"
int main()
{
std::string str = "Somewhere down the range v3 library";
for (auto s : str | ranges::view::split(' ')
| ranges::view::transform([](auto&& sub) { return std::string_view(&*sub.begin(), ranges::distance(sub)); }
))
{
std::cout << "Substring: " << s << "\n";
}
}
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, ';')
void splitString(string str, char delim, string array[], const int arraySize)
{
int delimPosition, subStrSize, subStrStart = 0;
for (int index = 0; delimPosition != -1; index++)
{
delimPosition = str.find(delim, subStrStart);
subStrSize = delimPosition - subStrStart;
array[index] = str.substr(subStrStart, subStrSize);
subStrStart =+ (delimPosition + 1);
}
}
STL还没有这样的方法。
但是,您可以通过使用std::string::C_str()成员来使用C的strtok()函数,也可以编写自己的函数。下面是我在快速谷歌搜索(“STL字符串分割”)后找到的代码示例:
void Tokenize(const string& str,
vector<string>& tokens,
const string& delimiters = " ")
{
// Skip delimiters at beginning.
string::size_type lastPos = str.find_first_not_of(delimiters, 0);
// Find first "non-delimiter".
string::size_type pos = str.find_first_of(delimiters, lastPos);
while (string::npos != pos || string::npos != lastPos)
{
// Found a token, add it to the vector.
tokens.push_back(str.substr(lastPos, pos - lastPos));
// Skip delimiters. Note the "not_of"
lastPos = str.find_first_not_of(delimiters, pos);
// Find next "non-delimiter"
pos = str.find_first_of(delimiters, lastPos);
}
}
摘自:http://oopweb.com/CPP/Documents/CPPHOWTO/Volume/C++编程-HOWTO-7.html
如果您对代码示例有疑问,请留下评论,我会解释。
仅仅因为它没有实现称为迭代器的typedef或重载<<运算符,并不意味着它是错误的代码。我经常使用C函数。例如,printf和scanf都比std::cin和std::cout快(很明显),fopen语法对二进制类型更友好,它们也倾向于生成更小的EXE。
不要被这种“优雅胜过性能”的交易所吸引。
值得一提的是,这里有另一种从输入字符串中提取令牌的方法,仅依赖于标准库设施。这是STL设计背后力量和优雅的一个例子。
#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <iterator>
int main() {
using namespace std;
string sentence = "And I feel fine...";
istringstream iss(sentence);
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
ostream_iterator<string>(cout, "\n"));
}
可以使用相同的通用复制算法将提取的令牌插入到容器中,而不是将其复制到输出流中。
vector<string> tokens;
copy(istream_iterator<string>(iss),
istream_iterator<string>(),
back_inserter(tokens));
…或直接创建矢量:
vector<string> tokens{istream_iterator<string>{iss},
istream_iterator<string>{}};
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!