CODEWITHSUNDEEP
cstringsequence
Oria Gruber
Oria Gruber

Reputation: 1533

Counting a sequence in a string C language

I'm trying to write a function that gets two strings, a short string called str1, and a longer one called str2.

the function will count how many times is str1 found in str2. 

example: str1= 'ab' str2 = 'abab' print 2

but for some reason, for the above example. i get 3. This is my code:

    int how_many_times(char* str1,char* str2)
{
    int result=0,i,j=0;
    for(i=0;i<strlen(str2);i++)
    {
        if((str1[i]==str2[j]))
        {
            while(str1[i]==str2[j])
            {
                j++;
                if(j==strlen(str1))
                {
                    result++;
                    j=0;
                }
                i++;
            }
        }
        else
            i++;
    }
    printf("%d",result);
    getch();
    return result;
}

Upvotes: 2

Views: 1337

Answers (4)

Marc Marta
Marc Marta

Reputation: 542

I came up with this (doesn't use strstr, but does use strlen):

int how_many_times(char *str, char *str2) {
    int result = 0, i = 0, j, k, len = strlen(str);

    while(str2[i]) {
        j = 0;
        k = 0;

        while(1) {
            if(str[j + k] == str2[i + k] && str[j] != 0) {
                if(j + k == len - 1) {
                    result++;
                    break;
                }
            } else {
                break;
            }
            k++;
        }

        i++;
    }

    return result;
}

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726699

Since this is likely a learning exercise, I would not suggest using a library function, assuming that you would rather fix your code.

Your code is close to working:

  • You swapped i and j in the code that accesses str[...] - you need to access str1 with j and str2 with i, not the other way around, because i goes up to strlen(str2) and j goes up to strlen(str1)
  • You need to reset j to zero when you find a mismatch - The else branch of the if needs to reset j after a partial match has been found.

Here is your modified code running on ideone.

int how_many_times(char* str1,char* str2)
{
    int result=0,i,j=0;
    for(i=0;i<strlen(str2);i++)
    {
        if((str1[j]==str2[i])) // <<<=== Swapped i and j
        {
            while(str1[j]==str2[i]) // <<<=== Swapped i and j
            {
                j++;
                if(j==strlen(str1))
                {
                    result++;
                    j=0;
                }
                i++;
            }
        }
        else {
            i++;
            j = 0; // <<<=== Added
        }
    }
    return result;
}

P.S. Library-based solution is better :-)

Upvotes: 2

Garee
Garee

Reputation: 1284

The strstr function returns a pointer to the first occurrence of a string within another. We can use this to count the number of occurrences by incrementing a variable count until the function returns NULL.

int countOccur(char *a, char *b) {
  int count = 0;
  while (b = strstr(b, a)) {
    count++;
    b++;
  }

  return count;
}

Upvotes: 1

ctn
ctn

Reputation: 2930

int count(char *s1, char *s2)
{
    char *p;
    int c;

    c = 0;
    for (p = strstr(s2, s1); p; p = strstr(p, s1)) {
        c++;
        p++;
    }
    return c;
}

Upvotes: 3

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