strstr 是否有反向函数

Question

我试图找到一个与strstr类似的函数，它搜索从字符串的末尾开始到开头的子字符串。

Answer 1

标准 C 库没有“反向 strstr”函数，因此您必须自己查找或编写。

我想出了几个自己的解决方案，并在该线程中添加了一些测试和基准测试代码以及​​其他功能。 对于那些好奇的人，在我的笔记本电脑（Ubuntu karmic，amd64 架构）上运行的输出如下所示：

$ gcc -O2 --std=c99 strrstr.c && ./a.out
#1 0.123 us last_strstr
#2 0.440 us theo
#3 0.460 us cordelia
#4 1.690 us digitalross
#5 7.700 us backwards_memcmp
#6 8.600 us sinan

您的结果可能会有所不同，并且根据您的编译器和库，结果的顺序也可能有所不同。

要从字符串的开头获取匹配项的偏移量（索引），请使用指针算法：

char *match = last_strstr(haystack, needle);
ptrdiff_t index;
if (match != NULL)
    index = match - haystack;
else
    index = -1;

现在，落叶松（请注意，这纯粹是用 C 语言编写的，我不太了解 C++，无法给出答案）：

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

/* By liw. */
static char *last_strstr(const char *haystack, const char *needle)
{
    if (*needle == '\0')
        return (char *) haystack;

    char *result = NULL;
    for (;;) {
        char *p = strstr(haystack, needle);
        if (p == NULL)
            break;
        result = p;
        haystack = p + 1;
    }

    return result;
}


/* By liw. */
static char *backwards_memcmp(const char *haystack, const char *needle)
{
    size_t haylen = strlen(haystack);

    if (*needle == '\0')
        return (char *) haystack;

    size_t needlelen = strlen(needle);
    if (needlelen > haylen)
        return NULL;

    const char *p = haystack + haylen - needlelen;
    for (;;) {
        if (memcmp(p, needle, needlelen) == 0)
            return (char *) p;
        if (p == haystack)
            return NULL;
        --p;
    }
}


/* From http://stuff.mit.edu/afs/sipb/user/cordelia/Diplomacy/mapit/strrstr.c
 */
static char *cordelia(const char *s1, const char *s2)
{
 const char *sc1, *sc2, *psc1, *ps1;

 if (*s2 == '\0')
  return((char *)s1);

 ps1 = s1 + strlen(s1);

 while(ps1 != s1) {
  --ps1;
  for (psc1 = ps1, sc2 = s2; ; )
   if (*(psc1++) != *(sc2++))
    break;
   else if (*sc2 == '\0')
    return ((char *)ps1);
 }
 return ((char *)NULL);
}


/* From http://stackoverflow.com/questions/1634359/
   is-there-a-reverse-fn-for-strstr/1634398#1634398 (DigitalRoss). */
static char *reverse(const char *s)
{
  if (s == NULL)
    return NULL;
  size_t i, len = strlen(s);
  char *r = malloc(len + 1);

  for(i = 0; i < len; ++i)
    r[i] = s[len - i - 1];
  r[len] = 0;
  return r;
}
char *digitalross(const char *s1, const char *s2)
{
  size_t  s1len = strlen(s1);
  size_t  s2len = strlen(s2);
  const char *s;

  if (s2len == 0)
    return (char *) s1;
  if (s2len > s1len)
    return NULL;
  for (s = s1 + s1len - s2len; s >= s1; --s)
    if (strncmp(s, s2, s2len) == 0)
      return (char *) s;
  return NULL;
}


/* From http://stackoverflow.com/questions/1634359/
  is-there-a-reverse-fn-for-strstr/1634487#1634487 (Sinan Ünür). */

char *sinan(const char *source, const char *target)
{
    const char *current;
    const char *found = NULL;

    if (*target == '\0')
        return (char *) source;

    size_t target_length = strlen(target);
    current = source + strlen(source) - target_length;

    while ( current >= source ) {
        if ( (found = strstr(current, target)) ) {
            break;
        }
        current -= 1;
    }

    return (char *) found;
}


/* From http://stackoverflow.com/questions/1634359/
  is-there-a-reverse-fn-for-strstr/1634441#1634441 (Theo Spears). */
char *theo(const char* haystack, const char* needle)
{
  size_t needle_length = strlen(needle);
  const char* haystack_end = haystack + strlen(haystack) - needle_length;
  const char* p;
  size_t i;

  if (*needle == '\0')
    return (char *) haystack;
  for(p = haystack_end; p >= haystack; --p)
  {
    for(i = 0; i < needle_length; ++i) {
      if(p[i] != needle[i])
        goto next;
    }
    return (char *) p;

    next:;
  }
  return 0;
}


/*
 * The rest of this code is a test and timing harness for the various
 * implementations above.
 */


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


/* Check that the given function works. */
static bool works(const char *name, char *(*func)(const char *, const char *))
{
    struct {
        const char *haystack;
        const char *needle;
        int offset;
    } tests[] = {
        { "", "", 0 },
        { "", "x", -1 },
        { "x", "", 0 },
        { "x", "x", 0 },
        { "xy", "x", 0 },
        { "xy", "y", 1 },
        { "xyx", "x", 2 },
        { "xyx", "y", 1 },
        { "xyx", "z", -1 },
        { "xyx", "", 0 },
    };
    const int num_tests = sizeof(tests) / sizeof(tests[0]);
    bool ok = true;

    for (int i = 0; i < num_tests; ++i) {
        int offset;
        char *p = func(tests[i].haystack, tests[i].needle);
        if (p == NULL)
            offset = -1;
        else
            offset = p - tests[i].haystack;
        if (offset != tests[i].offset) {
            fprintf(stderr, "FAIL %s, test %d: returned %d, haystack = '%s', "
                            "needle = '%s', correct return %d\n",
                            name, i, offset, tests[i].haystack, tests[i].needle,
                            tests[i].offset);
            ok = false;
        }
    }
    return ok;
}


/* Dummy function for calibrating the measurement loop. */
static char *dummy(const char *haystack, const char *needle)
{
    return NULL;
}


/* Measure how long it will take to call the given function with the
   given arguments the given number of times. Return clock ticks. */
static clock_t repeat(char *(*func)(const char *, const char *),
                       const char *haystack, const char *needle,
                       long num_times)
{
    clock_t start, end;

    start = clock();
    for (long i = 0; i < num_times; ++i) {
        func(haystack, needle);
    }
    end = clock();
    return end - start;
}


static clock_t min(clock_t a, clock_t b)
{
    if (a < b)
        return a;
    else
        return b;
}


/* Measure the time to execute one call of a function, and return the
   number of CPU clock ticks (see clock(3)). */
static double timeit(char *(*func)(const char *, const char *))
{
    /* The arguments for the functions to be measured. We deliberately
       choose a case where the haystack is large and the needle is in
       the middle, rather than at either end. Obviously, any test data
       will favor some implementations over others. This is the weakest
       part of the benchmark. */

    const char haystack[] = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "b"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
                            "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
    const char needle[] = "b";

    /* First we find out how many repeats we need to do to get a sufficiently
       long measurement time. These functions are so fast that measuring
       only a small number of repeats will give wrong results. However,
       we don't want to do a ridiculously long measurement, either, so 
       start with one repeat and multiply it by 10 until the total time is
       about 0.2 seconds. 

       Finally, we measure the dummy function the same number of times
       to get rid of the call overhead.

       */

    clock_t mintime = 0.2 * CLOCKS_PER_SEC;
    clock_t clocks;
    long repeats = 1;
    for (;;) {
        clocks = repeat(func, haystack, needle, repeats);
        if (clocks >= mintime)
            break;
        repeats *= 10;
    }

    clocks = min(clocks, repeat(func, haystack, needle, repeats));
    clocks = min(clocks, repeat(func, haystack, needle, repeats));

    clock_t dummy_clocks;

    dummy_clocks = repeat(dummy, haystack, needle, repeats);
    dummy_clocks = min(dummy_clocks, repeat(dummy, haystack, needle, repeats));
    dummy_clocks = min(dummy_clocks, repeat(dummy, haystack, needle, repeats));

    return (double) (clocks - dummy_clocks) / repeats / CLOCKS_PER_SEC;
}


/* Array of all functions. */
struct func {
    const char *name;
    char *(*func)(const char *, const char *);
    double secs;
} funcs[] = {
#define X(func) { #func, func, 0 }
    X(last_strstr),
    X(backwards_memcmp),
    X(cordelia),
    X(digitalross),
    X(sinan),
    X(theo),
#undef X
};
const int num_funcs = sizeof(funcs) / sizeof(funcs[0]);


/* Comparison function for qsort, comparing timings. */
int funcmp(const void *a, const void *b)
{
    const struct func *aa = a;
    const struct func *bb = b;

    if (aa->secs < bb->secs)
        return -1;
    else if (aa->secs > bb->secs)
        return 1;
    else
        return 0;
}


int main(void)
{

    bool ok = true;
    for (int i = 0; i < num_funcs; ++i) {
        if (!works(funcs[i].name, funcs[i].func)) {
            fprintf(stderr, "%s does not work\n", funcs[i].name);            
            ok = false;
        }
    }
    if (!ok)
        return EXIT_FAILURE;

    for (int i = 0; i < num_funcs; ++i)
        funcs[i].secs = timeit(funcs[i].func);
    qsort(funcs, num_funcs, sizeof(funcs[0]), funcmp);
    for (int i = 0; i < num_funcs; ++i)
        printf("#%d %.3f us %s\n", i+1, funcs[i].secs * 1e6, funcs[i].name);

    return 0;
}

Answer 2

我不知道一个。 C 的优点之一是，如果您编写自己的函数，它与库函数一样快速和高效。 （在许多其他语言中完全不是这种情况。）

您可以反转字符串和子字符串，然后进行搜索。

最后，当字符串库不够好时，人们经常做的另一件事是转向正则表达式。

好的，我写了reverse()和rstrstr() ，如果幸运的话，它们可能会起作用。 摆脱 C++ 的__restrict 。 您可能还希望将参数设为const ，但随后您将需要转换返回值。 要回答您的评论问题，您只需从中减去原始字符串指针即可从子字符串的地址中获取索引。 行：

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

char *reverse(const char * __restrict const s)
{
  if (s == NULL)
    return NULL;
  size_t i, len = strlen(s);
  char *r = malloc(len + 1);

  for(i = 0; i < len; ++i)
    r[i] = s[len - i - 1];
  r[len] = 0;
  return r;
}

char *rstrstr(char *__restrict s1, char *__restrict s2)
{
  size_t  s1len = strlen(s1);
  size_t  s2len = strlen(s2);
  char *s;

  if (s2len > s1len)
    return NULL;
  for (s = s1 + s1len - s2len; s >= s1; --s)
    if (strncmp(s, s2, s2len) == 0)
      return s;
  return NULL;
}

Answer 3

如果你可以使用 C++，你可以像这样搜索字符串：

std::string::iterator found=std::search(haystack.rbegin(), haystack.rend(), needle.rbegin(), needle.rend()).base();
// => yields haystack.begin() if not found, otherwise, an iterator past-the end of the occurence of needle

Answer 4

一种可能的（如果不是完全优雅的）实现可能如下所示：

#include "string.h"

const char* rstrstr(const char* haystack, const char* needle)
{
  int needle_length = strlen(needle);
  const char* haystack_end = haystack + strlen(haystack) - needle_length;
  const char* p;
  size_t i;

  for(p = haystack_end; p >= haystack; --p)
  {
    for(i = 0; i < needle_length; ++i) {
      if(p[i] != needle[i])
        goto next;
    }
    return p;

    next:;
  }
  return 0;
}

Answer 5

不。这是 C++ std::string 类具有明显优势的地方之一——除了std::string::find() ，还有std::string::rfind() 。

Answer 6

我认为你仍然可以使用库函数来做到这一点。

1.使用strrev函数反转字符串。

2.使用strstr函数做任何你想做的事。

3.您可以通过从原始字符串的长度中减去搜索字符串的起始索引来找到搜索字符串的起始索引（反向）。

Answer 7

尽管不是标准的，但strrstr得到了广泛的支持，并且完全符合您的要求。

Answer 8

是否有 C 库函数来查找字符串中子字符串最后一次出现的索引？

编辑：正如@hhafez 在下面的评论中指出的那样，我为此发布的第一个解决方案效率低下且不正确（因为我通过target_length提前了指针，这在我的愚蠢测试中运行良好）。 您可以在编辑历史记录中找到该版本。

这是一个从最后开始并返回的实现：

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

const char *
findlast(const char *source, const char *target) {
    const char *current;
    const char *found = NULL;

    size_t target_length = strlen(target);
    current = source + strlen(source) - target_length;

    while ( current >= source ) {
        if ( (found = strstr(current, target)) ) {
            break;
        }
        current -= 1;
    }

    return found;
}

int main(int argc, char *argv[]) {
    if ( argc != 3 ) {
        fputs("invoke with source and search strings as arguments", stderr);
        return EXIT_FAILURE;
    }

    const char *found = findlast(argv[1], argv[2]);

    if ( found ) {
        printf("Last occurence of '%s' in '%s' is at offset %d\n",
                argv[2], argv[1], found - argv[1]
                );
    }
    return 0;
}

输出：

C:\Temp> st "this is a test string that tests this" test
Last occurence of 'test' in 'this is a test string that tests this' is 
at offset 27

Answer 9

我不相信 c 字符串库中存在，但是编写自己的字符串会很简单，在一种情况下，您知道字符串的长度或者它已正确终止。

Answer 10

标准 C 库中没有。 您可以在网上找到一个，或者您可能需要自己编写。

Answer 11

长话短说：

不 - C 库中没有任何功能可以满足您的需求。

但正如其他人指出的那样：编写这样的函数不是火箭科学......

Answer 12

这是一个。 测试它是一个练习，我会留给你:)

Answer 13

感谢您的回答！ 还有一种来自 MSDN 论坛的方法。 http://social.msdn.microsoft.com/Forums/en-US/vclanguage/thread/ed0f6ef9-8911-4879-accb-b3c778a09d94

Answer 14

char * strrstr(char *_Str, char *_SubStr){
    char *returnPointer, *p;

    //find 1st occurence. if not found, return NULL
    if ( (p=strstr(_Str, _SubStr))==NULL)
        return NULL;

    //loop around until no more occurences
    do{
        returnPointer=p;
        ++p;
    }while(p=strstr(p, _SubStr));

    return returnPointer;
}

Answer 15

为此，您可以使用标准算法 std::find_end。 例如

    char s[] = "What is the last word last";
    char t[] = "last";

    std::cout << std::find_end( s, s + sizeof( s ) - 1, t, t + sizeof( t ) -1 )
              << std::endl;

Answer 16

这是我能想到的最小的简单植入。 与此函数的其他实现不同，它避免了像 user3119703 这样的其他人所拥有的初始 strstr 调用。

char * lastStrstr(const char * haystack,const char * needle){
    char*temp=haystack,*before=0;
    while(temp=strstr(temp,needle)) before=temp++;
    return before;
}

Answer 17

char* strrstr(char * _Str, const char * _SubStr)
{
    const BYTE EQUAL=0;
    int i=0, src_len = strlen(_Str), find_len = strlen(_SubStr),
        tail_count=0;

    for(i=src_len; i>-1; i--)
    {
        if(_Str[i] == _SubStr[0] && tail_count >= find_len)
        {
            if(strncmp(&_Str[i], _SubStr, find_len) == EQUAL)
            {
                return &_Str[i];
            }
        }
        tail_count++;
    }
    return NULL;    
}