Thread: anagram strings

Would someone help me please, it has to check if the second word is anagram to the first one.
So far ...

Code:

#include <stdio.h>

void main()
{


char first[20], second[20];
int x,y;
int firstSum =0;
int secondSum =0;

printf ("enter a word: ");
gets(first);

printf ("enter an anagram: ");
gets(second);

while(first[x] !='\0')
{   
    firstSum += first[x];
    x++;
}

while(second[y] !='\0')
{
    secondSum += second[y];
    y++;
}

if (firstSum==secondSum)
{
    printf(" is anagarm");
}

else
{
printf(" is not an anagarm");

}

}

A couple things first:
1. It's int main(void), and return an int at the end, usually zero for success. Read this: FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[]) - Cprogramming.com.
2. Don't use gets. Read this: FAQ > Why gets() is bad / Buffer Overflows - Cprogramming.com.

Now, to find an anagram, you don't just add up the numeric values of each character in the string. You have to count each letter independently. How many A's? How many B's? All the way down to Z.

Thank for the tips.

I'll think of counting each letter. I just want to know why this isn't working ?

My bad, I had to be more specific on my second question. The point is it crashes after I enter the second string into. I just want to know for now -why does it crashes?

Hm, x and y are uninitialized which means you don't know if the index is out of range, you are entering a string that is more than 20 characters. Two possible reasons.

x and y are uninitialized which means you don't know if the index is out of range

exactly , thank you

Don't think of counting each letter, do it. It's the only way*.

Since first and second are just arrays, it makes sense to use a for loop, which makes it harder to forget initialization:

Code:

for (x = 0; first[x] != '\0'; x++) {
    increment count of letter in first[x]
}

As for subsonics implementation in post #4, it has a bug. It will produce false positives if, say, s2 has two 'q's and s1 only has 1.

* Not really the only way, but it's probably the easiest and most intuitive way. There's also a method that involves sorting (case-insensitive) each string and comparing. Probably some other clever stuff too.

Yeah, that is true. What do you mean by counting letters? X keeps the count, are you referring to sum?

I meant count the occurrences of each character.

Originally Posted by anduril462

I meant count the occurrences of each character.

Ok I see. Here's an update using strchr btw.

Code:

int in_set(char *s1, char *s2) {
    for(char *p; *s2; s2++) {
        p = strchr(s1, *s2);
        if(p == NULL) {
            return 0;
        }
    }
    return 1;
}

int is_anagram(char *s1, char *s2) {
    if( strlen(s1) != strlen(s2) ) {
        return 0;
    }

    if(in_set(s1, s2)) {
        return in_set(s2, s1);
    }

    return 0;
}

Thanks! Now I've got to learn what is strchr

@subsonics what youre saying is true using pointer arithmetic to check if the letters of both words are the same make the anagram

@Subsonics: Your code still doesn't work. Try "foo" and "off" as your two words. They are not anagrams. Your program reports them as false positives.

Originally Posted by ph3s

Thanks! Now I've got to learn what is strchr

strchr man page