Char Arrays + Functions + For

Because buf is of type char[][] and word is of type char[].
So when you do buf[i], it becomes char[] and word[i] becomes char.
char[] != char.
Types mismatch.
It's like trying to assign a character to a string... but where in the string do you want to put that character?

Quote:

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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

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Oh no it won't. That's why I wanted to know what you are trying to do, because I think what you have done so far is pretty much beyond repair for any purpose. As laserlight, etc, have been trying to point out, this is because you are using a 2D array as if it were a 1D array, which implies you don't understand the difference.

Well.. but there I'm calling the first array of the 2d array, how can this assignment be done?
Yes, I understand the difference, that's why I was calling the first dimension of the array to be assigned

A simple drawing to show what you are doing:
buf looks like:
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]
[ ] [ ] [ ] [ ] [ ]

And word looks like:
[ ] [ ] [ ] [ ] [ ]

So now you choose a row in the buf, and you will get this:
[ ] [ ] [ ] [ ] [ ]

And then you choose an element in word, and you simply get one element:
[ ]

But where among the elements above do you wish to assign it? The compiler does not know.
buf[n]: [ ] [ ] [ ] [ ] [ ]

^
|
|

word[n]: [ ]

?

This is why it isn't working.

Quote:

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Originally Posted by lautarox

Well.. but there I'm calling the first array of the 2d array, how can this assignment be done?

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What do you mean by "I'm calling the first array of the 2d array"?

I understood it, I confused with the 2d array, thanks

If you're looking for how to find anagrams, this may help:
http://netlib.bell-labs.com/cm/cs/pearls/sketch02.html
http://netlib.bell-labs.com/cm/cs/pearls/s02.pdf

I've improved my code, but I cant count the letters correctly..
This is my output

Code:

---

tolook: daxear word: dear
Lenght: 4
Cantidad de buff[0] = 100
Cantidad de buff[1] = 101
Cantidad de buff[2] = 97
Cantidad de buff[3] = 114
Sec Lenght 6
Se econtro d en la pos: 0 de tolook
Cantidad de buff[0] en tolook 2
Se econtro e en la pos: 1 de tolook
Cantidad de buff[1] en tolook 2
Se econtro a en la pos: 2 de tolook
Se econtro a en la pos: 2 de tolook
Cantidad de buff[2] en tolook 2
Se econtro r en la pos: 3 de tolook
Cantidad de buff[3] en tolook 134514138
Char: d Cantidad: 134523305
Sec Char: d Cantidad: 2
Char: e Cantidad: -1076587899
Sec Char: e Cantidad: 2
Char: a Cantidad: -1209212939
Sec Char: a Cantidad: 2
Char: r Cantidad: -1208005407
Sec Char: r Cantidad: 134514138
No se puede

---

My code

Code:

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int sameWord(char *tolook, char *word){
        printf("tolook: %s word: %s\n", tolook, word);
        char lenght, seclenght, i,j;
        char buff[7];
        int cant[6];
        int seccant[6];
        char secbuff[7];
        lenght = strlen(word);
        printf("Lenght: %i\n", lenght);
        for(i=0;i<lenght;i++){
                buff[i] = word[i];
                for(j=0;j<lenght;j++){
                        if(buff[i] == word[j]){
                                cant[i]++;
                        }
                }
                printf("Cantidad de buff[%i] = %i\n", i, buff[i]);
        }
        seclenght = strlen(tolook);
        printf("Sec Lenght %i\n", seclenght);
        for(i=0;i<lenght;i++){
                for(j=0;j<seclenght;j++){
                        if(buff[i] == tolook[j]){
                                printf("Se econtro %c en la pos: %i de tolook\n", buff[i], i);
                                seccant[i]++;
                        }
                }
                printf("Cantidad de buff[%i] en tolook %i\n", i, seccant[i]);
        }
        j=0;
        int l;
        for(i=0;i<lenght;i++){
                printf("Char: %c Cantidad: %i\n", buff[i], cant[i]);
                printf("Sec Char: %c Cantidad: %i\n", buff[i], seccant[i]);
                if(cant[i] <= seccant[i]) {
                        j++;
                }
        }
        if(j==lenght) {
                return 1;
        }
        else {
                return 0;
        }
}

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Nice to see you are using printf to do some debugging!

(edit: sorry, now I was misreading you)



Anyway, I will keep reading thru it...

Quote:

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Originally Posted by MK27

I presume you mean to move a pointer forward. That would be cant++...

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No can do. "cant" is an array, not a pointer.

Just for you to have a better look:
Cantidad de buff[0] -> quantity of buff[0]
Se econtro d en la pos: 0 -> d was found in the 0 position
Cantidad de buff[3] en tolook -> quantity of buff[] in tolook

Sorry, I had to pause for food. But now let's look at your first for loop:

Code:

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for(i=0;i<lenght;i++){
                buff[i] = word[i];

---

So we are copying word into buff one letter at a time...

Code:

---

                for(j=0;j<lenght;j++){
                        if(buff[i] == word[j]){
                                    cant[i]++;
                        }

---

Now, we check to see if this letter is in word, and if so add one to cant[i]. Hmmm...

• you didn't initialize cant to anything, use int cant[6]={0} to make the whole array 0.
• since you copied from word to start with, this will always be true, so why not skip this inner loop and just say cant[i]++ ?

So the result of this is that word is copied into buff, and each element of cant is incremented by one. Let's move on to the second loop (skipping the printfs):

Code:

---

seclenght = strlen(tolook);
for(i=0;i<lenght;i++){
                for(j=0;j<seclenght;j++){
                        if(buff[i] == tolook[j]){
                                seccant[i]++;
                        }
                }
        }

---

Now we iterate through buff, checking each letter to see if it's in tolook. If so, the element of seccant (which is undefined again) corresponding to that letter is incremented by one.
Next:

Code:

---

        j=0;
        int l;        // what is this for and why is it here?
        for(i=0;i<lenght;i++){
                        if(cant[i] <= seccant[i]) {
                        j++;
                }
        }

---

First, j is reset to use as a counter in the loop, where we compare cant to seccant. Since (if the two arrays had been initialized to zero, that is) cant={1,1,1,1,1,1} and seccant's elements will be either zero or one, j will equal the number of letters from word that found a match in seccant. Then:

Code:

---

if(j==lenght) {
                return 1;
        }
        else {
                return 0;
        }

---

Wow, lottarocks, this might actually work if you had initialized your arrays to zero! But there is a lot of unnecessary stuff going on. How about this:

Code:

---

int cant[6]={0}, seccant[6]={0}, matches=0;
for(i=0;i<lenght;i++){                    // matches is new to replace j for counting
                buff[i] = word[i];
                cant[i]++;
}
seclenght = strlen(tolook);
for(i=0;i<lenght;i++){
                for(j=0;j<seclenght;j++){
                        if(buff[i] == tolook[j]){
                                matches++;
                        }
                }
        }
if(matches==lenght) {
                return 1;
        }
        else {
                return 0;
        }

---

If you look at this a little bit you should be able to see how to streamline things even further.

If I actually do this

Code:

---

for(i=0;i<lenght;i++){
                buff[i] = word[i];
                cant[i]++;

---

It won't count if there are two exact letters, that's why I'm doing a for and counting how many times it appears, I've initialized the arrays with 0 and I'm still having the same problem, look
tolook: daxear word: dear
Lenght: 4
Cantidad de buff[0] = 100 -> Why am I getting this high numers?
Cantidad de buff[1] = 101
Cantidad de buff[2] = 97
Cantidad de buff[3] = 114
Sec Lenght 6
Se econtro d en la pos: 0 de tolook
Cantidad de buff[0] en tolook 1
Se econtro e en la pos: 1 de tolook
Cantidad de buff[1] en tolook 1
Se econtro a en la pos: 2 de tolook <- Why is it doing it twice? and in the same position of tolook
Se econtro a en la pos: 2 de tolook
Cantidad de buff[2] en tolook 2
Se econtro r en la pos: 3 de tolook
Cantidad de buff[3] en tolook 1
Char: d Cantidad: 1
Sec Char: d Cantidad: 1
Char: e Cantidad: 1
Sec Char: e Cantidad: 1
Char: a Cantidad: 1
Sec Char: a Cantidad: 2
Char: r Cantidad: 1
Sec Char: r Cantidad: 1
Se Puede

Quote:

---

Originally Posted by lautarox

Cantidad de buff[0] = 100 -> Why am I getting this high numers?
Cantidad de buff[1] = 101
Cantidad de buff[2] = 97
Cantidad de buff[3] = 114

---

Hey great, these are actually letters:
100=d
101=e
97=a
114=r

google "codigos ascii" (or ascii table) and you will quickly figure it out. It's very important to understand with C and probably all computer languages.

lol I confused, I changed buff[i] for cant[i], now I realize..
But I can't fix the word counting &#172;&#172;