Thread: Char Arrays + Functions + For

Hint: word[i] is a char but buff[i] is not.

So, you can't assign an array with a character, right? And that's what your "line 30" tries to do.

--
Mats

Code:

		buff[i] = word[i]; -> First Error (line 30)

What is buff[i], and what is word[i]?

--
Mats

Originally Posted by lautarox

But I'm assigning it to the first value and so on..

What is "it", and what "first value" are you talking about?

Originally Posted by lautarox

I think it has to be a little problem..

I think it might be more than a little problem and if you followed laserlight's hint you might be starting to think the same thing!

Are you just trying to compare two words? If so, maybe start again...what's the purpose of buff[i][i]++? This will just change a letter to the next letter, so that if buff[i][i]='b', buff[i][i] will now be c. But at that point in the program flow, buff[i][i] was bound to be undefined anyway, so you will be taking a mystery value and adding 1 to it.

Also: the return type should be int, not char.

Well.. the purpose of my function is to see if a word can be found in 6 random letters, for example, "esters" the function takes a word and it has a buff array that stores the letter in the buff[i] and the times it appears in the buff[i][i], then it does the same with the other one and then checks if the word can be formed with the random letters

You still do not need a 2D char array for this (or two loops). There is something you are confused about.

This will do more or less what you are asking and might clarify some issues for you.

Code:

#include <stdio.h>
#include <string.h>

int sameWord(char *one, char *two) {  	
	int len1=strlen(one), len2=strlen(two), i,j;

	for (i=0;i<len1;i++) {
		for (j=0;j<len2;j++) {
			if (one[i]==two[j]) break;
			if (j==len2-1) return 0;           //end of "two", letter not found
		}
	}	
	return 1;
}

int main(int argc, char *argv[]) {
	if (argc<3) {puts("Two words needed");return -1;}
	if (sameWord(argv[1],argv[2])==1) printf("\"&#37;s\" has the letters to spell \"%s\"\n",argv[2],argv[1]);
	else printf("\"%s\" does not have the letters to spell \"%s\"\n",argv[2],argv[1]);
	return 0;
}

Output:

Code:

./a.out this xchtsi
"xchtsi" has the letters to spell "this"
./a.out this xcgtso
"xchgso" does not have the letters to spell "this"

Originally Posted by lautarox

But what if the word has 2 same letters?

So you don't want (?):
"xem" has the letters to spell "meme"

I suppose you could zero a letter in "two" out after it is used, so that it will never be found again (if you do that, you'll have to work with a copy of "two" or else it will be affected in main() after the function completes). But my point with the post was to demonstrate a (functioning) function with a for loop operating on char arrays, so that you could contrast this with what you were trying to do, and thereby get a better idea of what will work, vs. what you think will work, and hopefully bring the two closer together.

If you see what I mean. What you are trying to do is fairly simple, the reason you are having a hard time is because you don't have enough grasp of the C syntax & concepts involved. Which is a good reason to keep working at it. So you could try to do what I just suggested to the code I posted before. Practice, practice, practice!

Originally Posted by lautarox

Yes, sure, but why I'm having that error? why can't I assign that value to the buffer var? that will just help me a lot

If you are talking about the errors on lines 30 and 36 as per your original code example, the reason is that you cannot assign a char to an array, yet word[i] and tolook[i] are chars, and buff[i] and secbuff[i] are arrays.