c++ - 使用 C++11 正则表达式匹配文本范围

Question

我在 C++ 中尝试正则表达式，这里是一些代码

#include <iostream>
#include <regex>


int main (int argc, char *argv[]) {
  std::regex pattern("[a-z]+", std::regex_constants::icase);
  std::regex pattern2("excelsior", std::regex_constants::icase);
  std::string text = "EXCELSIOR";

  if (std::regex_match(text, pattern)) std::cout << "works" << std::endl;
  else std::cout << "doesn't work" << std::endl;

  if (std::regex_match(text, pattern2)) std::cout << "works" << std::endl;
  else std::cout << "doesn't work" << std::endl;

  return 0;
}

现在，据我了解，这两个匹配项都应该输出 works，但是第一个输出 doesn't work，而第二个输出 works 正如预期的那样。为什么？

Answer 1

根据[re.grammar]中描述的规则，我们有:

— During matching of a regular expression finite state machine against a sequence of characters, two characters c and d are compared using the following rules:
1. if (flags() & regex_constants::icase) the two characters are equal if traits_inst.translate_nocase(c) == traits_inst.translate_nocase(d);
2. otherwise, if flags() & regex_constants::collate the two characters are equal if traits_inst.translate(c) == traits_inst.translate(d);
3. otherwise, the two characters are equal if c == d.

这适用于您的 pattern2 ，我们正在匹配一个字符序列，我们有 flags() & icase , 所以我们做一个 nocase 比较。由于序列中的每个字符都匹配，因此它“有效”。

但是，对于 pattern ，我们没有字符序列。所以我们改用这个规则:

— During matching of a regular expression finite state machine against a sequence of characters, comparison of a collating element range c1-c2 against a character c is conducted as follows: if flags() & regex_constants::collate is false then the character c is matched if c1 <= c && c <= c2, otherwise c is matched in accordance with the following algorithm:

string_type str1 = string_type(1,
    flags() & icase ?
        traits_inst.translate_nocase(c1) : traits_inst.translate(c1);
string_type str2 = string_type(1,
    flags() & icase ?
        traits_inst.translate_nocase(c2) : traits_inst.translate(c2);
string_type str = string_type(1,
    flags() & icase ?
        traits_inst.translate_nocase(c) : traits_inst.translate(c);
return traits_inst.transform(str1.begin(), str1.end())
        <= traits_inst.transform(str.begin(), str.end())
    && traits_inst.transform(str.begin(), str.end())
        <= traits_inst.transform(str2.begin(), str2.end());

因为你没有 collate设置，字符在字面上与范围 a-z 匹配.没有会计 icase在这里，这就是它“不起作用”的原因。如果您提供 collate然而:

std::regex pattern("[a-z]+", 
                   std::regex_constants::icase | std::regex_constants::collate);

然后我们使用所描述的算法进行非大小写比较，结果将是“有效”。两个编译器都是正确的——尽管我发现在这种情况下预期的行为令人困惑。