问题描述

我在玩 Boost.Regex 来解析单词和数字的字符串.这是我目前所拥有的:

I was playing around with Boost.Regex to parse strings for words and numbers. This is what I have so far:

#include <iostream>
#include <string>
#include <boost/foreach.hpp>
#include <boost/regex.hpp>
#include <boost/range.hpp>

using namespace std;
using namespace boost;

int main()
{
    regex re
    (
        "("
            "([a-z]+)|"
            "(-?[0-9]+(\.[0-9]+)?)"
        ")"
    );

    string s = "here is a	 list of Words. and some 1239.32 numbers to 3323 parse.";
    sregex_iterator m1(s.begin(), s.end(), re), m2;

    BOOST_FOREACH (const match_results<string::const_iterator>& what, make_iterator_range(m1, m2)) {
        cout << ":" << what[1].str() << ":" << what.position(1) << ":" << what.length(1) << endl;
    }

    return 0;
}

有没有办法告诉正则表达式从流而不是字符串中解析?似乎应该可以使用任何迭代器.

Is there a way to tell regex to parse from a stream rather than a string? It seems like it should be possible to use any iterator.

推荐答案

Boost.IOStreams 有一个 regex_filter 允许在流上执行相当于 regex_replace 的操作.然而，从实现来看，它似乎是作弊"，因为它只是将整个流加载到缓冲区中，然后在该缓冲区上调用 Boost.Regex.

Boost.IOStreams has a regex_filter allowing one to perform the equivalent of a regex_replace on a stream. However, looking at the implementation, it seems to "cheat" in that it simply loads the whole stream into a buffer and then calls Boost.Regex on that buffer.

可以使用部分匹配" 支持 Boost.Regex.查看页面末尾的示例.

Making a regex search on a stream's contents without having to entirely load it in memory can be done with the "partial match" support of Boost.Regex. Look at the example at the end of the page.

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