first occurence in s1 compared to s2

**xion** · 09-09-2003

the following to pieces of code are functions i have attempted to write. its suppose to return the first location in s1 that any character in s2 occurs.

my question is why doesnt the first set of code work? i think its because of the 2nd for loop that has s2[j] != s1[i] but im not sure. the only differences between the first and a second sets of code is the extra variable and the "s2[j] !=s1[i]" part.

for those of you that have k&r this is ex. 2-5.
thanks in advance for response guys...and girls.

Code:

//non-working code

int any(char s1[], char s2[])
{
   int i,j;

   for(i = 0; s1[i] != '\0'; i++)
   {
      for(j = 0; s2[j] != '\0' && s2[j] != s1[i]; j++)
      {
         if (s1[i] == s2[j])
            return i;
         else
            return -1;
      }
   }
}

Code:

int any(char s1[], char s2[])
{
//working code
   int   i;
   int   j;
   int   pos;

   pos = -1;
   for (i = 0; s1[i] != '\0' && pos == -1; i++)
   {
      for (j = 0; s2[j] != '\0' && pos == -1; j++)
      {
         if (s2[j] == s1[i])
         {
            pos = i;
            return pos;
         }
      }
   }
}

i

**C-learning** · 09-09-2003

Code:

int any(char s1[], char s2[])
{
   int i,j;

   for(i = 0; s1[i] != '\0'; i++)
   {
      for(j = 0; s2[j] != '\0' && s2[j] != s1[i]; j++)
      {
         if (s1[i] == s2[j])
            return i;
         else
            return -1;
      }
   }
}

move return -1 outside both loops
This because you want to go thrue every character
in s1. Now your program only looks at the first character.
If s1[0] == s2[0] return 0 else return -1

**quzah** · 09-09-2003

Actually you're both wrong. If you are trying to get it to return the indices on any matching character, then you do not return when it isn't a match. Otherwise all that happens is it checks one single character. If that character matches, it returns 0, because that's the first indices, otherwise it returns -1.

You want to do nothing on a non-match, so that it continues through the loop.

If at the end you are outside of the loop, then you know that thee was no match, so then return -1.

Quzah.

**char** · 09-09-2003

Code:

int any(char s1[], char s2[])
{
    int i,j;

    for(i = 0; s1[i] != '\0'; i++)
    {
        for(j = 0; s2[j] != '\0' && s2[j] != s1[i]; j++)
            ;

        if (s1[i] == s2[j])
            return i;
    }

    return -1;
}

**C-learning** · 09-09-2003

Actually you're both wrong. If you are trying to get it to return the indices on any matching character, then you do not return when it isn't a match. Otherwise all that happens is it checks one single character. If that character matches, it returns 0, because that's the first indices, otherwise it returns -1.

Wasnt i actually saying this or is my english bad?

**quzah** · 09-10-2003

Originally posted by C-learning
Wasnt i actually saying this or is my english bad?

Yeah. You did. I just looked at the code you posted and assumed you were posting your version rather than quoting theirs. I just looked at the code rather than what you wrote. :P

Quzah.

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