c语言通过指针strstr

Question

is this the standard code for strstr i made????

char* fstrset(char *s,char *t)
{
    int b, i=0,j=0;

 while(*(s+i)!='\0')
 {
  if(*(t+j)=='\0')
   break;
  else if(*(s+i)==*(t+j))
   {
   i++;j++;b=1;
   }
  else
   { i++;b=0;j=0;
   }
 }

    if(b==0)
     return((char*)NULL);
    else if(b==1)
     return(s+i-j);
}

Answer 1

This is all the standard has to say about it:

7.21.5.7 The strstr function

Synopsis

 #include <string.h> char *strstr(const char *s1, const char *s2);

Description

The strstr function locates the first occurrence in the string pointed to by s1 of the sequence of characters (excluding the terminating null character) in the string pointed to by s2.

Returns

The strstr function returns a pointer to the located string, or a null pointer if the string is not found. If s2 points to a string with zero length, the function returns s1.

So, it looks like you're missing const qualifiers on arguments.

As for style, note that *(ptr+index) can be replaced by ptr[index] , and size_t is the best type to use for indexing a pointer.

As for being a common way to implement it, compare with GCC's code:

char *
strstr (const char *s1, const char *s2)
{
  const char *p = s1;
  const size_t len = strlen (s2);

  for (; (p = strchr (p, *s2)) != 0; p++)
    {
      if (strncmp (p, s2, len) == 0)
    return (char *)p;
    }
  return (0);
}

Answer 2

Your code is buggy. Given:

char *haystack = "fififi-trixabelle";
char *needle = "fifi-trixabelle";

fstrset(haystack, needle) returns incorrectly returns NULL .

Answer 3

Besides the bug mentioned by caf there are others:

1) Uninitialized b. If s points to '\\0' , closing brace may be reached, omitting any return statements.

2) If characters match up to the end of string pointed to by s there is no check if the string pointed to by t ends too.

Answer 4

What does this do? It looks like gibberish. Why adding pointers, and mixing them with ints? Sorry, but the whole thing doesn't make sense.

\n

And to answer your question, i don't think so. But if you compile it and it runs, then yes.

Okay, your code does make sense when you look at it closer. Yes, it does look like it will compile, if thats what you mean by standard code.

Answer 5

inline char* strstr(char* __s1, const char* __s2)
{
    return __builtin_strstr(const_cast<const char*>(__s1), __s2); 
}

Answer 6

a quick read through seems to show that the code works (there are probably edge cases that dont work). You tell us, does it work?

But why do it? just call strstr

Answer 7

There is no 'standard code', just the standard result.

It is unlikely that any implementation in a standard C library uses array indexing, so it is unlikely that your code matches any implementation in line-by-line detail.

Answer 8

char* fstrstr(char *s1,char *s2)
{
 int i=0,flag=0;
 char *s4,*s3;
// s4 for retaining the value of s2
 s4 = s2;
 while(*s1 != '\0' && *s2 != '\0')
 {
  if(*s1 == *s2)
  {
   *(s3+i) = *s1;
   s2++;
   s1++;
   i++;
   flag = 1;
  }
  else
  {
   i = 0;
   s1++;
//   Initialize s2 again from its address
   s2 = s4;
   flag = 0;
  }
 }
 if(flag == 1)
 {
  while(*s1 != '\0')
  {
   *(s3+i) = *s1;
   i++;
   s1++;
  }
  *(s3+i) = '\0';
 }
 if(flag == 1)
  return (s3);

 if(flag==0)
 {
  *s3 = NULL;
  return (s3);
 }
}

Answer 9

There is no "standard code", only standard results.

It is unlikely that any implementation in the standard C library will use array indexes, so your code is unlikely to match any implementation in the line implementation.

Answer 10

    char *strstr(const char *s1, const char *s2) {
      char *a = s1, *b = s2;
      for (;;)
        if      (!*b)          return (char *)s1;
        else if (!*a)          return NULL;
        else if (*a++ != *b++) {a = ++s1; b = s2;}
    }