perpos.cpp

// Le cout par position rejetee depend-il de la complexite du motif,
// ou est-ce un surcout fixe par tentative ?
#define NOMINMAX
#include <windows.h>
#include <chrono>
#include <cstdio>
#include <regex>
#include <string>

typedef std::chrono::steady_clock clk;
static const size_t N = 303104;
static std::string plain;

static void bench(const char* name, const char* pattern) {
    std::regex rx(pattern, std::regex::ECMAScript | std::regex::optimize);
    auto once = [&]() -> long {
        long n = 0;
        for (auto it = std::sregex_iterator(plain.begin(), plain.end(), rx);
             it != std::sregex_iterator(); ++it) ++n;
        return n;
    };
    long m = once();
    auto t0 = clk::now();
    int passes = 0;
    double el = 0;
    do { m = once(); ++passes; el = std::chrono::duration<double>(clk::now() - t0).count(); }
    while (el < 1.0);
    double ms = el / passes * 1000.0;
    printf("  %-34s %8.2f ms  %8.1f ns/position  %5ld matches\n",
           name, ms, ms * 1e6 / N, m);
}

int main() {
    plain.assign(N, 'x');
    for (size_t i = 0; i < N; ++i) plain[i] = (char)('a' + (i * 7 + i / 13) % 26);
    for (size_t i = 79; i < N; i += 80) plain[i] = '\n';

    printf("\n  std::regex, texte de %.0f Ko sans chiffre ni 'Z'\n\n", N / 1024.0);
    bench("motif complet (19 car., 6 groupes)",
          R"((\d{4})-(\d{2})-(\d{2})T(\d{2}):(\d{2}):(\d{2}))");
    bench("motif complet, sans quantificateur",
          R"((\d\d\d\d)-(\d\d)-(\d\d)T(\d\d):(\d\d):(\d\d))");
    bench("motif \\d{4}   (quantificateur)", R"(\d{4})");
    bench("motif \\d\\d\\d\\d (repete a la main)", R"(\d\d\d\d)");
    bench("motif \\d{1}   (quantificateur trivial)", R"(\d{1})");
    bench("motif \\d      (un seul noeud)", R"(\d)");
    bench("motif Z       (litteral)", R"(Z)");
    printf("\n");
    return 0;
}