在c++ 14中使用正则表达式处理字符串

Question

我需要从c++ 14中的字符串中提取3个变量。字符串格式为:

a single uppercase char + `->` + a single uppercase char + `,` + a number

例如:A->B,100,C->D,20000,E->F,22。我想提取单个大写字符和数字，如A,B,100。到目前为止，我可以编写一个tokenized函数，通过多次调用tokenized()将它们分开，如下所示:

vector<string> tokenize(const string& s, const char c)
{
vector<string> splitted;
auto ss = stringstream(s);
string tmp;
while(getline(ss, tmp, c)) { splitted.push_back(tmp); }
return splitted;
}
// ...
t = tokenized(s, '')

但我想知道是否有一种更简单的方法来提取这3个变量使用正则表达式?我试过(A-Z.*?->A-Z),，但显然我写错了正则表达式。

Answer 1

对于c++中的正则表达式，您需要知道一件事:您需要自己对多个匹配进行迭代。下面是一个例子:

#include <iostream>
#include <regex>
#include <string>
#include <vector>
//-----------------------------------------------------------------------------
// if you're going to tokenize you might as well return tokens
struct token_t
{
token_t(const std::string& f, const std::string& t, const unsigned long v) :
from{ f },
to{ t },
value{ v }
{
}
std::string from;
std::string to;
unsigned long value;
};
std::ostream& operator<<(std::ostream& os, const token_t& token)
{
os << "token : from = " << token.from << ", to = " << token.to << ", value = " << token.value << std::endl;
return os;
}
std::ostream& operator<<(std::ostream& os, const std::vector<token_t>& tokens)
{
std::cout << std::endl << "------------------ tokens ------------------" << std::endl;
for (const auto& token : tokens)
{
os << token;
}
std::cout << "--------------------------------------------" << std::endl;
return os;
}
//-----------------------------------------------------------------------------
auto tokenize(const std::string& s)
{
static const std::regex rx{ "([A-Z])->([A-Z]),([0-9]+)" };
std::smatch match;
std::vector<token_t> tokens;
auto from = s.cbegin();
while (std::regex_search(from, s.cend(), match, rx))
{
tokens.push_back({ match[1], match[2], std::stoul(match[3]) });
from = match.suffix().first;
}
return tokens;
}
//-----------------------------------------------------------------------------
int main()
{
auto v1 = tokenize("A->B,100");
auto v2 = tokenize("A->B,100, C->D,2000, E->F,22");
std::cout << v1;
std::cout << v2;
return 0;
}

Answer 2

您想要的模式是:

[A-Z]->[A-Z],[0-9]+

解释:

[A-Z]   single uppercase character
->      ->
[A-Z]   single uppercase character
,       ,
[0-9]+  whole number

注意[A-Z]和[0-9]是字符类，表示特定字符范围内的任何大写字符或数字。