Question

我正在开发一个C / C ++应用程序（在Visual Studio 2010中），我需要对逗号分隔的字符串进行标记，我希望它尽可能快。目前我正在使用strtok_s。我运行了一些strtok_s与sscanf的测试，似乎strtok_s更快（除非我写了一个糟糕的实现:)）但我想知道是否有人可以提出更快的替代方案。< / p>

Answer 1

对于纯粹的运行时速度，boost。spirit。qi是一个很好的候选人。

Answer 2

您可以做的最好的事情是确保只通过字符串一次，并动态构建输出。开始将字符拉入临时缓冲区，当遇到分隔符时将临时缓冲区保存到输出集合，清除临时缓冲区，冲洗并重复。

这是执行此操作的基本实现。

template<class C=char>
struct basic_token
{
    typedef std::basic_string<C> token_string;
    typedef unsigned long size_type;
    token_string token_, delim_;
    basic_token(const token_string& token, const token_string& delim = token_string());
};

template<class C>
basic_token<C>::basic_token(const token_string& token, const token_string& delim)
:   token_(token),
    delim_(delim)
{
}

typedef basic_token<char> token;

template<class Char, class Iter> void tokenize(const std::basic_string<Char>& line, const Char* delims, Iter itx)
{
    typedef basic_token<Char> Token;
    typedef std::basic_string<Char> TString;

    for( TString::size_type tok_begin = 0, tok_end = line.find_first_of(delims, tok_begin);
        tok_begin != TString::npos; tok_end = line.find_first_of(delims, tok_begin) )
    {
        if( tok_end == TString::npos )
        {
            (*itx++) = Token(TString(&line[tok_begin]));
            tok_begin = tok_end;
        }
        else
        {
            (*itx++) = Token(TString(&line[tok_begin], &line[tok_end]), TString(1, line[tok_end]));
            tok_begin = tok_end + 1;
        }
    }
}

template<class Char, class Iter> void tokenize(const Char* line, const Char* delim, Iter itx)
{
    tokenize(std::basic_string<Char>(line), delim, itx);
}
template<class Stream, class Token> Stream& operator<<(Stream& os, const Token& tok)
{
    os << tok.token_ << "\t[" << tok.delim_ << "]";
    return os;
}

...你可以这样使用：

string raw = "35=BW|49=TEST|1346=REQ22|1355=2|1182=88500|1183=88505|10=087^";
vector<stoken> tokens;
tokenize(raw, "|", back_inserter(tokens));
copy(tokens.begin(), tokens.end(), ostream_iterator<stoken>(cout, "\n"));

输出是：

35=BW   [|]
49=TEST [|]
1346=REQ22      [|]
1355=2  [|]
1182=88500      [|]
1183=88505      [|]
10=087^ []

Answer 3

我想提醒你，strtok及其同类存在风险你可以获得不同数量的令牌。

one|two|three  would yield 3 tokens

，而

one|||three    would yield 2.

Answer 4

mmhmm的测试没有正确使用精神，他的语法是缺陷。

#include <cstdio> 
#include <cstring>   

#include <iostream>
#include <string>

#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>

#include <boost/spirit/include/qi.hpp>    

/****************************strtok_r************************/
typedef struct sTokenDataC {
    char *time;
    char *symb;
    float bid;
    float ask;
    int bidSize;
    int askSize;
} tokenDataC;

tokenDataC parseTick( char *line, char *parseBuffer )
{
    tokenDataC tokenDataOut;

    tokenDataOut.time = strtok_r( line,",", &parseBuffer );
    tokenDataOut.symb = strtok_r( nullptr,",", &parseBuffer );
    tokenDataOut.bid = atof(strtok_r( nullptr,",", &parseBuffer ));
    tokenDataOut.ask = atof(strtok_r( nullptr , ",", &parseBuffer ));
    tokenDataOut.bidSize = atoi(strtok_r( nullptr,",", &parseBuffer ));
    tokenDataOut.askSize = atoi(strtok_r( nullptr, ",", &parseBuffer  ));

    return tokenDataOut;
}

void test_strcpy_s(int iteration)
{
    char *testStringC = new char[64];
    char *lineBuffer = new char[64];

    printf("test_strcpy_s....\n");
    strcpy(testStringC,"09:30:00,TEST,13.24,15.32,10,14");
    {
        timeEstimate<> es;
        tokenDataC tokenData2;
        for(int i = 0; i < iteration; i++)
        {
            strcpy(lineBuffer, testStringC);//this is more realistic since this has to happen because I want to preserve the line
            tokenData2 = parseTick(lineBuffer, testStringC);
            //std::cout<<*tokenData2.time<<", "<<*tokenData2.symb<<",";
            //std::cout<<tokenData2.bid<<", "<<tokenData2.ask<<", "<<tokenData2.bidSize<<", "<<tokenData2.askSize<<std::endl;
        }
    }

    delete[] lineBuffer;
    delete[] testStringC;
}
/****************************strtok_r************************/

/****************************spirit::qi*********************/
namespace qi = boost::spirit::qi;

struct tokenDataCPP
{
    std::string time;
    std::string symb;
    float bid;
    float ask;
    int bidSize;
    int askSize;

    void clearTimeSymb(){
        time.clear();
        symb.clear();
    }
};

BOOST_FUSION_ADAPT_STRUCT(
        tokenDataCPP,
        (std::string, time)
        (std::string, symb)
        (float, bid)
        (float, ask)
        (int, bidSize)
        (int, askSize)
        )

void test_spirit_qi(int iteration)
{
    std::string const strs("09:30:00,TEST,13.24,15.32,10,14");
    tokenDataCPP data;        

    auto myString = *~qi::char_(",");
    auto parser = myString >> "," >> myString >> "," >> qi::float_ >> "," >> qi::float_ >> "," >> qi::int_  >> "," >> qi::int_;
    {
        std::cout<<("test_spirit_qi....\n");
        timeEstimate<> es;
        for(int i = 0; i < iteration; ++i){
            qi::parse(std::begin(strs), std::end(strs), parser, data);
            //std::cout<<data.time<<", "<<data.symb<<", ";
            //std::cout<<data.bid<<", "<<data.ask<<", "<<data.bidSize<<", "<<data.askSize<<std::endl;
            data.clearTimeSymb();
        }
    }
}
/****************************spirit::qi*********************/

int main()
{
    int const ITERATIONS = 500 * 10000;
    test_strcpy_s(ITERATIONS);
    test_spirit_qi(ITERATIONS);
}

由于clang ++没有strtok_s，我使用strtok_r来替换它迭代500 * 10k，时间是

test_strcpy_s：1.40951
test_spirit_qi：1.34277

他们的时间差不多，差别不大。

编译器，clang ++ 3.2，-O2

代码timeEstime

Answer 5

这应该非常快，没有临时缓冲区，它也分配空的代币。

$date = 2010-04-19;

$newDate = date("Y-m",strtotime($date));

$stringWithDate = "onlinearticles_".$newDate;

dd($stringWithDate);

Answer 6

在对每个建议的候选人进行测试和计时后，结果显示strtok显然是最快的。虽然我对测试的热爱并不感到惊讶，但是值得探索其他选择。 [注意：代码被编辑在一起编辑欢迎:)]

假设：

typedef struct sTokenDataC {
    char *time;
    char *symb; 
    float bid;
    float ask;
    int bidSize;
    int askSize;
} tokenDataC;

tokenDataC parseTick( char *line, char *parseBuffer )
{
    tokenDataC tokenDataOut;

    tokenDataOut.time = strtok_s( line,",", &parseBuffer );
    tokenDataOut.symb = strtok_s( null,",", &parseBuffer );
    tokenDataOut.bid = atof(strtok_s( null,",", &parseBuffer ));
    tokenDataOut.ask = atof(strtok_s( null , ",", &parseBuffer ));
    tokenDataOut.bidSize = atoi(strtok_s( null,",", &parseBuffer ));
    tokenDataOut.askSize = atoi(strtok_s( null, ",", &parseBuffer  ));

    return tokenDataOut; 
}

char *testStringC = new char[64];
    strcpy(testStringC,"09:30:00,TEST,13.24,15.32,10,14");

int _tmain(int argc, _TCHAR* argv[])
{
char *lineBuffer = new char[64];
    printf("Testing method2....\n");
    for(int i = 0; i < ITERATIONS; i++)
    {
        strcpy(lineBuffer,testStringC);//this is more realistic since this has to happen because I want to preserve the line
        tokenData2 = parseTick(lineBuffer,parseBuffer);

    }
}

vs通过以下方式致电John Diblings：

    struct sTokenDataCPP
    {
        std::basic_string<char> time;
        std::basic_string<char> symb; 
        float bid;
        float ask;
        int bidSize;
        int askSize;
    };
        std::vector<myToken> tokens1;
            tokenDataCPP tokenData;
            printf("Testing method1....\n");
            for(int i = 0; i < ITERATIONS; i++)
            {
tokens1.clear();
                tokenize(raw, ",", std::back_inserter(tokens1));

                tokenData.time.assign(tokens1.at(0).token_);
                tokenData.symb.assign(tokens1.at(1).token_);
                tokenData.ask = atof(tokens1.at(2).token_.c_str());
                tokenData.bid = atof(tokens1.at(3).token_.c_str());
                tokenData.askSize = atoi(tokens1.at(4).token_.c_str());
                tokenData.bidSize = atoi(tokens1.at(5).token_.c_str());

            }

vs一个简单的boost.spirit.qi实现，定义语法如下：

template <typename Iterator>
struct tick_parser : grammar<Iterator, tokenDataCPP(), boost::spirit::ascii::space_type>
{

    tick_parser() : tick_parser::base_type(start)
    {
        my_string %= lexeme[+(boost::spirit::ascii::char_ ) ];

        start %=
            my_string >> ','
            >>  my_string >> ','
            >>  float_ >> ','
            >>  float_ >> ','
            >>  int_ >> ','
            >>  int_
            ;
    }

    rule<Iterator, std::string(), boost::spirit::ascii::space_type> my_string;
    rule<Iterator, sTokenDataCPP(), boost::spirit::ascii::space_type> start;
};

将ITERATIONS设置为500k： strtok版本：2s 约翰的版本：115s 提升：172s

我可以发布完整的代码是人们想要这个，我只是不想占用巨大的空间

C / C ++中的快速字符串标记化

6 个答案: