在字符串中查找子字符串

Question

I am writing a function to find substring. But I am not sure where i am going wrong. On running GDB i get a segmentation fault. If someone can guide me in the right direction.

here is the code

char *mystrstr(char * s1, const char * s2)



int main(){
    char *s1 = "The quick brown fox jumps over the hell lazy dog";
    char *s2 = "hello";
    char *s4;
    s4 = mystrstr(s1,s2);
    printf("%s\n",s4);  <--- this is where i am Seg. Faulting



    return 0;
}

Answer 1

When s2 is not a substring of s1 you are returning null, and then you are trying to print it, that gives a segmentation fault.

Try something like this:

s4 = mystrstr(s1,s2);
if(s4 != NULL)
    printf("%s\n",s4);

Answer 2

The problem is that in the inner loop you add the indices i+j to access s1 . If you imagine i to point to the "o" in "dog" in your example, j goes from 0 to 5 (length of "hello") in the inner loop. This causes your access to s1[i+j] to look at the characters o , g , \\0 , garbage , garbage .

The benefit of C strings is that they are null terminated. So you can iterate over strings like

for (char* i = s1; *i != 0; i++) {
    ...
}

Ie you iterate from the start of s1 until you find its terminating 0 byte. In your inner loop, this allows you to write the following:

const char *j, *k;
for (j = s2, k = i; *j == *k && *j != 0; j++, k++);
if (*j == 0)
    return i;

Ie j starts at the beginning of s2 , k starts where i is currently pointing at inside s1 . You iterate as long as both strings are equal, and they have not reached their terminating 0 byte. If you have indeed reached the 0 byte of s2 ( *j == 0 ), you have found the substring.

Note that you probably want to return i instead of s1 , since this gives you a pointer into s1 where the requested substring starts.

Answer 3

printf("%s\n",s4? s4 : "(NULL)");

Answer 4

char *s1 = "The quick brown fox jumps over the hell lazy dog";
char *s2 = "hello";

According to mystrstr *s2 is not substring of *s1 because you cannot find hello in the *s1 . Therefore, the method will return NULL . And printing NULL as a string is not possible and result in an error.

To verify this, try to replace *s1 by:

char *s1 = "The quick brown fox jumps over the hello lazy dog"; // replace hell by hello

The output will be:

The quick brown fox jumps over the hello lazy dog