与通配符匹配的文件名

Question

我需要实现类似我自己的文件系统的东西。一个操作是 FindFirstFile。我需要检查调用者是否传递了类似 .、sample*.cpp 等内容。我的“文件系统”实现提供“文件名”列表作为 char* 数组。

Windows有没有函数或源代码实现这种文件名匹配？

Answer 1

对于使用“*”和“?”的通配符名称匹配试试这个（如果你想避免提升，请使用 std::tr1::regex）：

#include <boost/regex.hpp>
#include <boost/algorithm/string/replace.hpp>

using std::string;

bool MatchTextWithWildcards(const string &text, string wildcardPattern, bool caseSensitive /*= true*/)
{
    // Escape all regex special chars
    EscapeRegex(wildcardPattern);

    // Convert chars '*?' back to their regex equivalents
    boost::replace_all(wildcardPattern, "\\?", ".");
    boost::replace_all(wildcardPattern, "\\*", ".*");

    boost::regex pattern(wildcardPattern, caseSensitive ? regex::normal : regex::icase);

    return regex_match(text, pattern);
}

void EscapeRegex(string ®ex)
{
    boost::replace_all(regex, "\\", "\\\\");
    boost::replace_all(regex, "^", "\\^");
    boost::replace_all(regex, ".", "\\.");
    boost::replace_all(regex, "$", "\\$");
    boost::replace_all(regex, "|", "\\|");
    boost::replace_all(regex, "(", "\\(");
    boost::replace_all(regex, ")", "\\)");
    boost::replace_all(regex, "{", "\\{");
    boost::replace_all(regex, "{", "\\}");
    boost::replace_all(regex, "[", "\\[");
    boost::replace_all(regex, "]", "\\]");
    boost::replace_all(regex, "*", "\\*");
    boost::replace_all(regex, "+", "\\+");
    boost::replace_all(regex, "?", "\\?");
    boost::replace_all(regex, "/", "\\/");
}

Answer 2

周围有很多这样的功能。这是各种实现的目录，分为递归和非递归，如果

您不喜欢那里的许可（或者链接有问题等），这里有一种可能的匹配算法实现，该算法至少非常接近 Windows 使用的算法：

#include <string.h>
#include <iostream>

bool match(char const *needle, char const *haystack) {
    for (; *needle != '\0'; ++needle) {
        switch (*needle) {
        case '?': 
            if (*haystack == '\0')
                return false;
            ++haystack;
            break;
        case '*': {
            if (needle[1] == '\0')
                return true;
            size_t max = strlen(haystack);
            for (size_t i = 0; i < max; i++)
                if (match(needle + 1, haystack + i))
                    return true;
            return false;
        }
        default:
            if (*haystack != *needle)
                return false;
            ++haystack;
        }
    }
    return *haystack == '\0';
}

#ifdef TEST
#define CATCH_CONFIG_MAIN

#include "catch.hpp"

TEST_CASE("Matching", "[match]") {
    REQUIRE(match("a", "a") == true);
    REQUIRE(match("a", "b") == false);
    REQUIRE(match("a*", "a") == true);
    REQUIRE(match("a?", "a") == false);
    REQUIRE(match("a?", "ab") == true);
    REQUIRE(match("a*b", "ab") == true);
    REQUIRE(match("a*b", "acb") == true);
    REQUIRE(match("a*b", "abc") == false);
    REQUIRE(match("*a*??????a?????????a???????????????", 
        "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa") == true);
}

#endif

由于讨论了某些算法的复杂性其他答案，我会注意到我相信这具有 O(NM) 复杂性和 O(M) 存储使用（其中 N 是目标字符串的大小，M 是模式的大小）。

使用 @masterxilo 的测试对：

"*a*??????*a*?????????a???????????????", "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"

...这在我的机器上大约 3 微秒内找到了匹配项。这比典型模式慢很多——我的大多数其他测试在这台特定机器上运行大约 300 纳秒左右。

同时，@masterxilo 的代码在同一台机器上运行大约需要 11 微秒，因此速度仍然快了大约 3 到 4 倍（更不用说更小、更简单了）。

Answer 3

查看 POSIX 函数 fnmatch、glob 和 wordexp。

Answer 4

这是我的尝试。

它是“C++”，但我故意保持它几乎完全与 C 兼容。
为了将其转换为 C，您所需要做的就是删除 template 部分并将 Pattern 和 Text 更改为类似 字符常量*。

// TEST THIS before use! I've only done limited testing.

#include <stddef.h>
#include <stdlib.h>
#include <string.h>

template<class Pattern, class Text>
bool wildcard(
    Pattern const pat_begin, Pattern const pat_end,
    Text text_begin, Text const text_end)
{
    ptrdiff_t const pat_size = pat_end - pat_begin;
    ptrdiff_t stackbuf[64];
    size_t c = sizeof(stackbuf) / sizeof(*stackbuf);
    ptrdiff_t *p = stackbuf;
    size_t n = 0;
    p[n++] = 0;
    while (n > 0 && text_begin != text_end)
    {
        for (size_t i = 0; i < n; i++)
        {
            if (p[i] == pat_size)
            {
                p[i--] = p[--n];
                continue;
            }
            switch (*(pat_begin + p[i]))
            {
            case '?': ++p[i]; break;
            case '*':
                ptrdiff_t off;
                off = p[i];
                while (off < pat_size &&
                    *(pat_begin + off) == '*')
                { ++off; }
                if (n == c)
                {
                    ptrdiff_t const *const old = p;
                    c *= 2;
                    if (c == 0) { ++c; }
                    size_t const size = c * sizeof(*p);
                    p = (ptrdiff_t *)realloc(
                        old == stackbuf ? NULL : p,
                        size);
                    if (old == stackbuf)
                    { memcpy(p, old, n * sizeof(*old)); }
                }
                p[n++] = off;
                break;
            default:
                if (*(pat_begin + p[i]) == *text_begin)
                { ++p[i]; }
                else { p[i--] = p[--n]; }
                break;
            }
        }
        ++text_begin;
    }
    bool success = false;
    if (text_begin == text_end)
    {
        while (!success && n > 0)
        {
            --n;
            while (p[n] != pat_size &&
                *(pat_begin + p[n]) == '*')
            { ++p[n]; }
            if (p[n] == pat_size)
            { success = true; }
        }
    }
    if (p != stackbuf) { free(p); }
    return success;
}

bool wildcard(char const *const pattern, char const *const text)
{
    return wildcard(
        pattern,
        pattern + (pattern ? strlen(pattern) : 0),
        text,
        text + (text ? strlen(text) : 0));
}

bool wildcard(wchar_t const *const pattern, wchar_t const *const text)
{
    return wildcard(
        pattern,
        pattern + (pattern ? wcslen(pattern) : 0),
        text,
        text + (text ? wcslen(text) : 0));
}

当然，您可以随意以任何您想要的方式使用该代码。 :)

Answer 5

Mehrdad 提供的解决方案具有指数级的运行时间，因为它使用回溯。花了一秒钟才弄清楚 "*a*??????*a*?????????a??????????????? ?”,“啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊aaa" 是一个匹配。此外，也无法将字符串与“*”或“?”进行匹配。在其中。

这是我提出的 O(nm) 实现，其内存使用量为 O(n)，其中 n 是表达式的长度，m 是字符串的长度。它还支持转义 ?、* 和 \。

#include <stddef.h>
#include <stdlib.h>
#include <string.h>

/** 
 * Wildcard matching.
 * See for example
 * http://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/find_c_search_wildcard.mspx?mfr=true
 *
 *  * matches any amount of any characters, 
 *  ? matches any single character.
 *  c matches c.
 *  No escaping of * and ? implemented (these are not allowed in windows filenames anyways).
 *
 * Backtracking O(2^n) implementation.
 *
 * from http://stackoverflow.com/questions/3300419/file-name-matching-with-wildcard 
 * http://stackoverflow.com/a/12231681/524504
 */
template<class Pattern, class Text>
bool wildcard(
        Pattern const pat_begin,
        Pattern const pat_end,
        Text text_begin, 
        Text const text_end)
{
    ptrdiff_t const pat_size = pat_end - pat_begin;

    /* initial pattern position stack (offsets into the pattern string) and its size c */
    ptrdiff_t stackbuf[64];
    size_t c                     = sizeof(stackbuf) / sizeof(*stackbuf);
    /* Stack base. Can be realloc'ed to some new memory location if we need more */
    ptrdiff_t *p                 = stackbuf;

    /* pointer to the location in the stack */
    size_t n                     = 0;
    p[n++]                       = 0; /* position 0 in the stack not used */

    /* text_begin updated to skip everything successfully consumed  */
    while (n > 0 && text_begin != text_end) 
    {
        for (size_t i = 0; i < n; i++) 
        {
            /* if we're at the end of the pattern, but not at the end of the 
             * string, we might have done something wrong.
             */
            if (p[i] == pat_size) 
            {
                p[i--] = p[--n];
                continue;
            }
            /* current pattern character */
            switch (*(pat_begin + p[i]))
            {
                case '?': ++p[i]; break; /* simply advance pattern pointer */
                case '*':
                          ptrdiff_t off;
                          off = p[i];
                          while (off < pat_size &&
                                  *(pat_begin + off) == '*')
                          { ++off; }
                          /* if the stack is full, reallocate */
                          if (n == c)
                          {
                              ptrdiff_t const *const old = p;
                              c *= 2;
                              /* assert positive size
                               * stack size never reduced anyways?
                               */
                              if (c == 0) { ++c; }
                              size_t const size = c * sizeof(*p);
                              /* cannot use realloc to copy original stack 
                               * (ptr in realloc must be 
                               * "Pointer to a memory block previously 
                               * allocated with malloc, calloc or realloc."
                               * must do manually
                               */
                              p = (ptrdiff_t *)realloc(
                                       old == stackbuf ? NULL : p,
                                      size);
                              if (old == stackbuf)
                              { memcpy(p, old, n * sizeof(*old)); }
                          }
                          /* store offset */
                          p[n++] = off;
                          break;
                default: /* a normal character in the pattern */
                          /* must be matched exactly */
                          if (*(pat_begin + p[i]) == *text_begin)
                          { ++p[i]; } /* advance pattern pointer */
                          else /* if not, backtrack */
                          { p[i--] = p[--n]; }
                          break;
            }
        }
        ++text_begin;
    }

    bool success = false;
    if (text_begin == text_end)
    {
        while (!success && n > 0)
        {
            --n;
            while (p[n] != pat_size &&
                    *(pat_begin + p[n]) == '*')
            { ++p[n]; }
            if (p[n] == pat_size)
            { success = true; }
        }
    }

    /* need to free stack if it was reallocated */
    if (p != stackbuf) { free(p); }
    return success;
}

bool wildcard(char const *const pattern, char const *const text)
{
    return wildcard(
            pattern,
            pattern + (pattern ? strlen(pattern) : 0),
            text,
            text + (text ? strlen(text) : 0));
}

bool wildcard(wchar_t const *const pattern, wchar_t const *const text)
{
    return wildcard(
            pattern,
            pattern + (pattern ? wcslen(pattern) : 0),
            text,
            text + (text ? wcslen(text) : 0));
}

/**
 * Virtual Machine Style Regular Expression parsing/NFA emulation
 * used for wildcard matching. O(nm) algorithm.
 *
 * See http://swtch.com/~rsc/regexp/ for more on efficient RegEx parsing.
 *
 * Copyright (c) March 29, 2013 Paul Frischknecht
 * Can be distributed under the MIT licence, see bottom of file.
 */

//#define wildcard_fast_DEBUG /* define to make the alogorithm printf what its doing */

/**
 * Instructions are:
 *
 * star
 *   This launches a new thread at the current position and continues with the 
 *   current thread at the next position accepting any character.
 * char c
 *   Accepts exactly the character c
 * anychar
 *   Accepts any character.
 * end 
 *   Accepts the end of the input string.
 */
enum InstructionType {
    End     = 0,
    Star    = 1,
    Char    = 2,
    AnyChar = 3
};

struct Instruction {
    InstructionType i;
    int c;       /* the caracter this instruction matches - undefined for non Char */
    int threads; /* threads active in the given instruction */
    /*
     * storing this here allows us to find out wether there is a thread
     * active at a given instruction in O(1)
     */
};

/** 
 * Wildcard (file path) matching.
 * See for example 
 * http://www.microsoft.com/resources/documentation/windows/xp/all/proddocs/en-us/find_c_search_wildcard.mspx?mfr=true
 *
 *  * matches any amount of any characters, 
 *  ? matches any single character.
 *  c matches c.
 *
 * Escaping of '*', '?' and '\' via '\*', '\?' and '\\' implemented. 
 * All other bytes are recognized directly as is. 
 * The string may not end in a single (uneven amount of) '\'.
 *
 * If text_end is 0, the text is assumed to be 0 terminated.
 * Same for pattern_end. The end is exclusive.
 *
 * @return A pointer to the character after the last character matched.
 *
 * Virtual machine O(nm) implementation, see http://swtch.com/~rsc/regexp/regexp2.html
 *
 * TODO Factor out the precompilation step to speed up matching multiple strings
 * against the same expression.
 */
template<class Pattern, class Text>
Text wildcard_fast(
        Pattern const pat_begin,
        Pattern const pat_end,
        Text const text_begin, 
        Text const text_end)
{
    /* 
     * give some reasonable default program size so we don't have to call
     * malloc in most cases 
     */
    #define DEFAULTonstack_program_SIZE 256
    Instruction onstack_program[DEFAULTonstack_program_SIZE];
    /* this stores the current run and next run thread list, alternating */
    /* there are as most as many threads as instructions */
    Instruction* onstack_threads[DEFAULTonstack_program_SIZE*2]; 

    Instruction** threads      = onstack_threads;
    Instruction*  program      = onstack_program;
    int           program_size = sizeof(onstack_program)/sizeof(*program);

    Instruction* program_last_instruction = program + program_size - 1;

    /* program and pattern pointers */
    Instruction* pp   = program;
    Pattern      patp = pat_begin;

    /* compile */

    while ((pat_end == 0 && *patp != 0) || (pat_end != 0 && patp != pat_end)) {

        /* need more space */
        if (pp == program_last_instruction) {

            Instruction*  old_program = program;
            Instruction** old_threads = threads;
            int old_program_size      = program_size;

            program_size *= 2;

            program = (Instruction*) malloc(program_size*sizeof(*program));
            threads = (Instruction**)malloc(program_size*sizeof(*threads)*2);

            memcpy(program, old_program, old_program_size*sizeof(*program));

            if (old_program != onstack_program) {
                free(old_program); free(old_threads);
            }

            program_last_instruction = program + program_size - 1;
            pp = pp - old_program + program;
        }

        /* parse pattern */
        switch (*patp) {
            case '*': 
                pp->i = Star; 
                /* Optimize multiple stars away */
                while ((pat_end == 0 || patp+1 != pat_end) && *(patp+1) == '*')  
                    patp++; 
                break;

            case '?':
                pp->i = AnyChar; 
                break;

            case '\\': 
                pp->i = Char; 
                pp->c = *(++patp); /* assumes string does not end in \ */
                break;

            default: 
                pp->i = Char; 
                pp->c = *patp; 
                break;
        }

        pp->threads = 0;

        pp++;
        patp++;
    }

    /* add the End instruction at the end */
    program_last_instruction = pp;
    pp->i                    = End;
    pp->threads              = 0;

    /* run */
    Text sp = text_begin; /* input string pointer */
    int n = 1, c = 0; /* next and current index */
    int threadcount[2];

    /* initialize */
    threadcount[c] = 1;
    threads[0] = program;
    threads[0]->threads++;

    /* run over text */
    while ((text_end == 0 && *sp != 0) || (text_end != 0 && sp != text_end)) {
        /* unless recreated, all threads will die */
        threadcount[n] = 0;

        /* run over threads */
        for (int i = 0; i < threadcount[c]; i++) {

            Instruction* inst = threads[2*i+c];
            switch (inst->i) {
                case End: 
                    /* we may not reach end early */ 
                    /* kill this thread without recrating it */
                    inst->threads--; 
                    continue; /* with for loop */

                case Char: 
                    if (*sp != inst->c) {
                        /* if the character is not matched, kill this thread */
                        inst->threads--;
                        continue;
                    }
                    break;

                case Star: 
                    /* spawn off additional thread at current location */

                    if (inst->threads == 1) { 
                        /* only if there's noone active there yet */
                        threads[2*(threadcount[n]++)+n] = inst;
                        inst->threads++;
                    }
                    break;
            }
            /* 
             * common actions: increase program counter for current thread,
             * decrese amount of threads in last (current) instruction.
             */
            inst->threads--;
            inst++;
            inst->threads++;

            /* respawn at new location in next iteration */
            threads[2*(threadcount[n]++)+n] = inst;

            if (inst->i == Star && (inst+1)->threads == 0) { 
                /* 
                 * already follow no-match option to give us 
                 * more stuff to do 
                 */
                threads[2*(threadcount[n]++)+n] = inst+1;
                (inst+1)->threads++;
            }

#ifdef wildcard_fast_DEBUG
              for (int i = 0 ; i < threadcount[n]; i++) {
                printf("thread %d at %d.\n", i, threads[2*i+n]-program);
            }
#endif

        }

#ifdef wildcard_fast_DEBUG
        const char *ns[] = {
            "end",
            "star",
            "char",
            "anychar",
        };
        for (Instruction* p = program; p->i; p++) {
            printf("%d. %s %c (%d threads)\n", p-program, ns[p->i], p->i == Char ? p->c : ' ', p->threads);
        }
#endif

        /* swap next and current and advance */
        n = c;
        c = !c;
        sp++;
    }

    /* 
     * if there is no thread active in the End instruction when the 
     * end of the input was reached, this was no match
     */
    if (program_last_instruction->threads == 0) sp = 0; 

    if (program != onstack_program) {
        /* only need to free if we used malloc */
        free(program); free(threads);
    }

    return sp;
}

char const* wildcard_fast(
        char const *const pattern, 
        char const *const text) 
{
    return wildcard_fast(
            pattern,
            (const char*)0,
            text,
            (const char*)0);
}

wchar_t const* wildcard_fast(
        wchar_t const *const pattern,
        wchar_t const *const text) 
{
    return wildcard_fast(
            pattern,
            (const wchar_t*)0,
            text,
            (const wchar_t*)0);
}


/* tests */
#ifndef INCLUDE_IMPLEMENTATION
/* 
 * I just include this file in my projects and define this.
 * That works well for simple algorithms like this
 */

#include <stdio.h>
#include <time.h>

int main() {
    struct {
        char* p;
        char* t;
        bool expected_result;
    } test[] = {
        {
            "",
            "",
            true
        },
        {
            "a",
            "",
            false
        },
        {
            "",
            "a",
            false
        },
        {
            "a",
            "a",
            true
        },        
        {
            "****.txt", 
            "hello.txt",
            true
        },
        {
            "*.txt", 
            "hello.tzt",
            false
        },
        {
            "*.t?t*", 
            "hello.tzt",
            true
        },
        {
            "*.*", 
            "hi.there",
            true
        },
        /* the wildcard implementation will fail this as it doesn't understand escaping */
        {
            "\\*who\\?\\?.*\\\\", 
            "*who??.there\\",
            true
        },        
        /* these take extraordinaryly long on the O(2^n) implementation */
        {
            "**a*************************??????***a************************?????????a???????????????", 
            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
            true
        },
        /* runs a bit better if we omit the extra *'s. The fast implementation already optimizes these away. */
        {
            "*a*??????*a*?????????a???????????????", 
            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
            true
        },
        {0,0} /* marks end of tests */
    };

    int t0 = clock();
    const char* result;
    const char* r[] = {"no match", "match"};

    for (int i = 0; test[i].p; i++) {
        printf("=== Test %d: Matching '%s' against '%s'...===\n", i, test[i].t, test[i].p);

        /* wildcard_fast */
        t0 = clock();
        result = wildcard_fast(test[i].p, test[i].t);

        printf("  wildcard_fast (took %d ms):\n", clock()-t0);
        if (!!result != test[i].expected_result)
            printf("    Test failed (reported %s instead of expected %s).\n",
                    r[!!result], r[test[i].expected_result]);
        else if (result) {
            printf("    %s\n", test[i].t);
            printf("    %*.c\n", result-test[i].t+1, '^');
        }
        else
            printf("    No match.\n");

        /* wildcard */
        t0 = clock();
        result = (const char*)
            wildcard(test[i].p, test[i].t);

        printf("  wildcard (took %d ms):\n", clock()-t0);
        if (!!result != test[i].expected_result)
            printf("    Test failed (reported %s instead of expected %s).\n",
                    r[!!result], r[test[i].expected_result]);
        else if (result) printf("    Match.\n");
        else printf("    No match.\n");

        printf("\n");
    }
}
#endif

/*
 * Permission is hereby granted, free of charge, to any person
 * obtaining a copy of this software and associated
 * documentation files (the "Software"), to deal in the
 * Software without restriction, including without limitation
 * the rights to use, copy, modify, merge, publish, distribute,
 * sublicense, and/or sell copies of the Software, and to
 * permit persons to whom the Software is furnished to do so,
 * subject to the following conditions:
 * 
 * The above copyright notice and this permission notice shall
 * be included in all copies or substantial portions of the
 * Software.
 * 
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY
 * KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE
 * WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR
 * PURPOSE AND NONINFRINGEMENT.  IN NO EVENT SHALL THE AUTHORS
 * OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR
 * OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
 * OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE
 * SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
 */

这使用了虚拟机方法。有关高效正则表达式解析的更多信息，请参阅 http://swtch.com/~rsc/regexp/ 。

Answer 6

这是一个无依赖的可移植 C++ 版本：

#include <string>

#include <string.h>

bool wild_match(const std::string& str, const std::string& pat) {
  std::string::const_iterator str_it = str.begin();
  for (std::string::const_iterator pat_it = pat.begin(); pat_it != pat.end();
       ++pat_it) {
    switch (*pat_it) {
      case '?':
        if (str_it == str.end()) {
          return false;
        }

        ++str_it;
        break;
      case '*': {
        if (pat_it + 1 == pat.end()) {
          return true;
        }

        const size_t max = strlen(&*str_it);
        for (size_t i = 0; i < max; ++i) {
          if (wild_match(&*(pat_it + 1), &*(str_it + i))) {
            return true;
          }
        }

        return false;
      }
      default:
        if (*str_it != *pat_it) {
          return false;
        }

        ++str_it;
    }
  }

  return str_it == str.end();
}

Answer 7

PathMatchSpec。尽管它受到 MAX_PATH 限制（即可接受不超过 260 个字符）。你最好实现自己的匹配器；这不是很多代码。

Answer 8

这与 @nabulke 的答案几乎相同，使用 C++11 而不是 Boost （所以如果您喜欢这个答案，请务必给这个答案点赞）：

#include <regex>
#include <string>

std::regex wildcardToRegex(const std::string& wildcard, bool caseSensitive = true)
{
    // Note It is possible to automate checking if filesystem is case sensitive or not (e.g. by performing a test first time this function is ran)
    std::string regexString{ wildcard };
    // Escape all regex special chars:
    regexString = std::regex_replace(regexString, std::regex("\\\\"), "\\\\");
    regexString = std::regex_replace(regexString, std::regex("\\^"), "\\^");
    regexString = std::regex_replace(regexString, std::regex("\\."), "\\.");
    regexString = std::regex_replace(regexString, std::regex("\\$"), "\\$");
    regexString = std::regex_replace(regexString, std::regex("\\|"), "\\|");
    regexString = std::regex_replace(regexString, std::regex("\\("), "\\(");
    regexString = std::regex_replace(regexString, std::regex("\\)"), "\\)");
    regexString = std::regex_replace(regexString, std::regex("\\{"), "\\{");
    regexString = std::regex_replace(regexString, std::regex("\\{"), "\\}");
    regexString = std::regex_replace(regexString, std::regex("\\["), "\\[");
    regexString = std::regex_replace(regexString, std::regex("\\]"), "\\]");
    regexString = std::regex_replace(regexString, std::regex("\\+"), "\\+");
    regexString = std::regex_replace(regexString, std::regex("\\/"), "\\/");
    // Convert wildcard specific chars '*?' to their regex equivalents:
    regexString = std::regex_replace(regexString, std::regex("\\?"), ".");
    regexString = std::regex_replace(regexString, std::regex("\\*"), ".*");

    return std::regex(regexString, caseSensitive ? std::regex_constants::ECMAScript : std::regex_constants::icase);
}

bool wildmatch(const std::string& input, const std::string& wildcard)
{
    auto rgx = wildcardToRegex(wildcard);
    return std::regex_match(input, rgx);
}

Answer 9

您也可以尝试我的通配符匹配库： Wildest card

它是：

• 基于
• 用 C 编写的 NFA在单个头文件中
• 没有依赖项
• 没有递归
• 没有动态分配
• 复杂度为 O(MN)
• 需要 O(logM) 内存

示例：

#include "wildcard.h"
...
bool answer = wildcard("*.cpp", "file.cpp");

Answer 10

这是一个高效的方法，它对字符串进行一次遍历，并在不需要的地方跳过比较（模式中的最后一个字符恰好是“*”，或者如果“*”之前的字符没有相同的字符）索引匹配）。它也不会通过使用指针来复制任何字符串，并且不会浪费任何 CPU 周期来使用 strlen()、sizeof()、.length() 或 .size() 获取字符串长度

#include <stdio.h>

_Bool wildcard_strcmp(char *line, char *pattern)
{
    _Bool wildcard = 0;
    char *placeholder;

    do
    {
        if ((*pattern == *line) || (*pattern == '?'))
        {
            line++;
            pattern++;
        }
        else if (*pattern == '*')
        {
            if (*(++pattern) == '\0')
                return 1;
            else
            {
                placeholder = pattern;
                wildcard = 1;
            }
        }
        else if (wildcard)
        {
            if (pattern == placeholder)
                line++;
            else
                pattern = placeholder;
        } 
        else
            return 0;
    } while (*line);

    if (*pattern == '\0')
        return 1;
    else
        return 0;
}

int main()
{
    char string[200] = "foobarfoobar";
    char pattern[200] = "fo?*barfoo*";

    if (wildcard_strcmp(string, pattern))
        printf("Match\n");
    else
        printf("No Match\n");

    return 0;
}