我想解析一个可以包含“-”的字符串，但不能以它开头或结尾。
我希望这个解析器能够工作：
auto const parser = alnum >> -(*(alnum | char_('-')) >> alnum);
但是在我的测试输入“某事”中，它只解析“所以”而不吃剩下的。
麻烦的是中间位*(alnum | char_('-'))一直吃到最后（包括最后一个字符，所以整个可选括号都失败了）。
我想知道的是，我怎样才能绕过它并制作这个解析器？
问问题
143 次
2 回答
1
我个人会“积极”地写它：
auto const rule = raw [ lexeme [
alnum >> *('-' >> alnum | alnum) >> !(alnum|'-')
] ];
这使用
lexeme处理空白重要性，raw避免必须主动匹配您想要作为输出一部分的每个字符（您只需要所有字符）。'-' >> alnum积极要求任何破折号后跟一个 alnum。"--"请注意，这在输入中也是非法的。请参阅下面的变体
#include <boost/spirit/home/x3.hpp>
#include <iostream>
#include <string>
#include <algorithm>
namespace x3 = boost::spirit::x3;
namespace parser {
using namespace boost::spirit::x3;
auto const rule = raw [ lexeme [
alnum >> *('-' >> alnum | alnum) >> !(alnum|'-')
] ];
}
int main() {
struct test { std::string input; bool expected; };
for (auto const t : {
test { "some-where", true },
test { " some-where", true },
test { "some-where ", true },
test { "s", true },
test { " s", true },
test { "s ", true },
test { "-", false },
test { " -", false },
test { "- ", false },
test { "some-", false },
test { " some-", false },
test { "some- ", false },
test { "some--where", false },
test { " some--where", false },
test { "some--where ", false },
})
{
std::string output;
bool ok = x3::phrase_parse(t.input.begin(), t.input.end(), parser::rule, x3::space, output);
if (ok != t.expected)
std::cout << "FAILURE: '" << t.input << "'\t" << std::boolalpha << ok << "\t'" << output << "'\n";
}
}
变体
为了也允许some--thing和类似的输入，我将更'-'改为+lit('-')：
alnum >> *(+lit('-') >> alnum | alnum) >> !(alnum|'-')
#include <boost/spirit/home/x3.hpp>
#include <iostream>
#include <string>
#include <algorithm>
namespace x3 = boost::spirit::x3;
namespace parser {
using namespace boost::spirit::x3;
auto const rule = raw [ lexeme [
alnum >> *(+lit('-') >> alnum | alnum) >> !(alnum|'-')
] ];
}
int main() {
struct test { std::string input; bool expected; };
for (auto const t : {
test { "some-where", true },
test { " some-where", true },
test { "some-where ", true },
test { "s", true },
test { " s", true },
test { "s ", true },
test { "-", false },
test { " -", false },
test { "- ", false },
test { "some-", false },
test { " some-", false },
test { "some- ", false },
test { "some--where", true },
test { " some--where", true },
test { "some--where ", true },
})
{
std::string output;
bool ok = x3::phrase_parse(t.input.begin(), t.input.end(), parser::rule, x3::space, output);
if (ok != t.expected)
std::cout << "FAILURE: '" << t.input << "'\t" << std::boolalpha << ok << "\t'" << output << "'\n";
}
}
于 2016-08-04T09:49:33.350 回答
0
我通过告诉贪婪的kleene星内的解析器忽略'eoi'（输入结束）来修复它。一个更强大的修复也会让它因空白而失败：
所以*(alnum | char_('-'))变成*((alnum | char_('-')) >> !(eoi | space))
于 2016-07-25T08:44:56.043 回答