C ++ TR1正则表达式 - 多行选项 [英] C++ TR1 regex - multiline option

问题描述

我认为$表示字符串的结束。然而，下面的代码段给出了testbbbccc作为结果，这对我来说是非常令人惊讶的...这意味着$实际匹配行结束，而不是整个字符串的结尾。

I thought that $ indicates the end of string. However, the following piece of code gives "testbbbccc" as a result, which is quite astonishing to me... This means that $ actually matches end of line, not end of the whole string.

#include <iostream>
#include <regex>

using namespace std;

int main()
{
    tr1::regex r("aaa([^]*?)(ogr|$)");
    string test("bbbaaatestbbbccc\nddd");
    vector<int> captures;
    captures.push_back(1);
    const std::tr1::sregex_token_iterator end;
    for (std::tr1::sregex_token_iterator iter(test.begin(), test.end(), r, captures); iter != end; )
    {
        string& t1 = iter->str();
        iter++;
        cout &lt;&lt; t1;
    }
} 

我一直在试图找到一个多线 （实际上可以在PCRE中轻松找到），但没有成功...有人可以指向正确的方向吗？

I have been trying to find a "multiline" switch (which actually can be easily found in PCRE), but without success... Can someone point me to the right direction?

注意，
RP

Regards, R.P.

推荐答案

由于tr1选择了Boost :: Regex，请尝试以下操作：

As Boost::Regex was selected for tr1, try the following:

从 Boost :: Regex

锚定：

Anchors:

一个行的开始
用作表达式的第一个
字符或
子表达式的第一个字符。

A '^' character shall match the start of a line when used as the first character of an expression, or the first character of a sub-expression.

当用作表达式的最后一个字符
或子表达式的最后一个
字符时，'$'字符将匹配
a行的结尾。

A '$' character shall match the end of a line when used as the last character of an expression, or the last character of a sub-expression.

因此，您观察到的行为是正确的。

So the behavior you observed is correct.

From： Boost Regex ：

\\ A 在缓冲区开始处匹配
（与 \\`）。

\\ z 在
匹配仅缓冲区的结尾与
\\'相同。

\\ Z 匹配可选序列
在缓冲区结尾处的换行符：
相当于正则表达式
\\ n * \\ z

\A Matches at the start of a buffer only (the same as \`).
\z Matches at the end of a buffer only (the same as \').
\Z Matches an optional sequence of newlines at the end of a buffer: equivalent to the regular expression \n*\z

我希望有帮助。

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