Here I go again...arrays....char...

My code is here! Vowels from sentence in one array, all none vowels in other array. First one is working but I can't get it to put all others in other array! Help!

Code:

---

int main()

{
        int sranje;
        char niz[100];
        char niz2[100]={NULL};
        char niz4[100]={NULL};
        char niz3[]={'a','A','e','E','U','u','i','I','o','O'};
        int a,b,c=0,d=0,br=0;

        printf("Upisi niz: ");
        fgets(niz, sizeof(niz), stdin);

for(a=0;a<strlen(niz)+1;a++)
{
                if(niz[a]==' ')
                        br++;
        for(b=0;b<strlen(niz3)+1;b++)
        {
                if(niz[a]==niz3[b])
                {
                        niz2[c]=niz[a];
                        c++;
                }
        }
}
                puts(niz2);
                printf("\n");
                puts(niz4);
                printf("\n%d", br);

        scanf("%d", &sranje);
}

---

I tried with combinations like this,

Code:

---

                          f(niz[a]!=niz3[b])
                        {
                                niz4[d]=niz[a];
                                d++;
                        }

---

but, nothing.

strlen(niz3)

to use this your niz3 should be null-terminated.
In your case - it is not. It is just a char array, so no string manipulation routines can be applied.
Or use sizeof(niz3)
Or initialize it in another way - to get the null-termination character. For example

Code:

---

char niz3[]="aAeEUuiIoO";

---

I did and if I implement this

Code:

---

if(niz[a]!=niz3[b])
                        {
                                niz4[d]=niz[a];
                                d++;
                        }

---

after first if, and input is "aaa", then it returns for "niz4" - aaaaaaaaaaaaaaaaaaaa

Adding one to the results of strlen is basically saying "explicit buffer overrun here please". Then leaving off the nul terminator char, well you just really really want to crash something don't you? :devil:

I advise single-stepping the whole way through your program.

Well the a proper solution is to test every letter of the input buffer to see whether is a vowel or not. Each time add every letter to its category and get the final results.
I think that a simple answer code is the following, think of bugs included.

Code:

---

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

const char *Vowls = "aeiou";

int GetVowels(const char *Buf, char **Vowels, char **NoVowels)
{
        if(Buf == NULL)
        {
                printf("Empty Buffer.\n");
                return -1;
        }
        else
        {
                //Variables.
                int i;
                int j = 0;
                int n = 0;
                //Get memory for the pointers.
                *Vowels = calloc(1, sizeof(char));
                *NoVowels = calloc(1, sizeof(char));
                //Check for memory fault.
                if(*Vowels == NULL || *NoVowels == NULL)
                {
                        printf("Memory Error.\n");
                        return -2;
                }
                else
                {
                        //Search each letter seperately to find what you need, save it to char arrays.
                        for(i = 0; i < (int)strlen(Buf); i++)
                        {
                                //Check for valid input, you can do whatever you want with it, i prefer to break as an invalid input.
                                if(Buf[i] == ';' || Buf[i] == ',' || Buf[i] == '!' || Buf[i] == ':')
                                        break;
                                if(strchr(Vowls, Buf[i]) != NULL)
                                {
                                        Vowels[0] = realloc(Vowels[0], sizeof(char) * (j+1));
                                        *(Vowels[0]+ j) = Buf[i];
                                        j++;
                                }
                                else
                                {
                                        NoVowels[0] = realloc(NoVowels[0], sizeof(char) * (n+1));
                                        *(NoVowels[0]+ n) = Buf[i];
                                        n++;
                                }
                        }
                        if(i == strlen(Buf))
                        {
                                //Get space for the '\0'.
                                Vowels[0] = realloc(Vowels[0], sizeof(char) * (j+1));
                                NoVowels[0] = realloc(NoVowels[0], sizeof(char) * (n+1));
                                //Put it.
                                *(Vowels[0]+ j) = '\0';
                                *(NoVowels[0]+ n) = '\0';
                                return 0;
                        }
                        else
                        {
                                printf("Internal Loop Error, wrong misplaced symbol &#37;c in Buffer in place %d.\n", Buf[i], i);
                                free(*Vowels);
                                free(*NoVowels);
                                *Vowels = NULL;
                                *NoVowels = NULL;
                                return -3;
                        }
                }
        }
}
int main(int argc, char *argv[])
{
        char Buf[256] = "";
        char *nv = NULL;
        char *v = NULL;
        printf("Give me the sentance:");
        fgets(Buf, sizeof(Buf), stdin);
        if(Buf[strlen(Buf) - 1] == '\n')
                Buf[strlen(Buf) - 1] = '\0';
        GetVowels(Buf, &v, &nv);
        printf("NoVowels: %s\n",nv);
        printf("Vowels:  %s\n",v);
        free(v);
        free(nv);
        return 0;
}

---

I dont remember exactly if the Vowels are these i declared but you can easily add or remove letters from the string before the function!!!

Some results:

Give me the sentance:Hi this is the new way of making code less friendly
NoVowels: H ths s th nw w f mkng cd lss frndl
Vowels: iiieeayoaioeeiey
Press any key to continue . . .

Bokarinho!

Bokarinho, that's an awful lot of energy to expend on a simple sieve. You sure like to be verbose! There exists a function in the standard C string library that can make such things a trivial exercise.

Quote:

---

Originally Posted by citizen

Bokarinho, that's an awful lot of energy to expend on a simple sieve. You sure like to be verbose! There exists a function in the standard C string library that can make such things a trivial exercise.

---

Seconded. That's a wicked amount of wheel reinvention there. That and 'y' is not a vowel.

Quote:

---

Originally Posted by iMalc

Seconded. That's a wicked amount of wheel reinvention there. That and 'y' is not a vowel.

---

Thanks for your great commentary.... Yes y is not a vowel you can easily remove it from the string and its all ok!!! Whats the problem with you people??????

Programmers are pedantic by nature

QuantumPete

Bokarinho

Quote:

---

I think that a simple answer code is the following, think of bugs included.

---

thx, but i think there is a much simpler way to do this, but i can learn something from your code, if i can't use that in this task ill use it in another.

Quote:

---

Adding one to the results of strlen is basically saying "explicit buffer overrun here please". Then leaving off the nul terminator char, well you just really really want to crash something don't you?

---

can you please be more specific. My new code looks like this now:

Code:

---

        char niz[100];
        char niz2[100]={NULL};
        char niz4[100]={NULL};
        char niz3[]="aAeEiIoOuU";
        int a,b,c=0,d=0,br=0;
        int i, j, temp=0;

        printf("Upisi niz: ");
        fgets(niz, sizeof(niz), stdin);

for(a=0;a<strlen(niz);a++)
{
                if(niz[a]==' ')
                {
                        br++;
                }
        for(b=0;b<strlen(niz3);b++)
        {
                if(niz[a]==niz3[b])
                {
                        niz2[c]=niz[a];
                        c++;
                }
                if(niz[a]!=niz3[b])
                {
                        niz4[d]=niz[a];
                        d++;
                }
               
        }
}

---

like i said before, vowels ok aa , novowels puts : aaaaaaaaaaaaaaaaaaaaaaaa

> like i said before, vowels ok aa , novowels puts : aaaaaaaaaaaaaaaaaaaaaaaa

You need to work ouy a way to check all of the vowels at once, given some letter in an input string. Right now, you're code only looks for 'a'. What you have simply doesn't work for the same reason that 'a' is not 'e' or any other letter.

Look up a function called strchr(), find out what it does, and think about how it can help you with your new problem. Or strcspn(), again.

Well, fu**'it, I can always do it like this, nice and sorted!!!

Code:

---

#include <stdio.h>
#include <string.h>
int main()
{
        int sranje;
        char niz[100];
        char niz2[100]={NULL};
        char niz4[100]={NULL};
        char niz3[]="aAeEiIoOuU";
        char niz5[]="bBcCdDfFgGhHkKlLpPwWrRtTzZQqsSyYxXjJVMmnN";
        int a,b,c=0,d=0,br=0;
        int i, j, temp=0;

        printf("Upisi niz: ");
        fgets(niz, sizeof(niz), stdin);

for(a=0;a<strlen(niz)-1;a++)
{
                if(niz[a]==' ')
                {
                        br++;
                }
        for(b=0;b<strlen(niz3)-1;b++)
        {
                if(niz[a]==niz3[b])
                {
                        niz2[c]=niz[a];
                        c++;
                }
        }
        for(b=0;b<strlen(niz5)-1;b++)
        {
                if(niz[a]==niz5[b])
                {
                        niz4[d]=niz[a];
                        d++;
                }
        }
}
for(c=0;c<strlen(niz2);c++)
{
        for(j=0;j<strlen(niz2);j++)
        {
                if(niz2[c]<niz2[j])
                {
                        temp=niz2[c];
                        niz2[c]=niz2[j];
                        niz2[j]=temp;
                }
        }
}
for(c=0;c<strlen(niz4);c++)
{
        for(j=0;j<strlen(niz4);j++)
        {
                if(niz4[c]<niz4[j])
                {
                        temp=niz4[c];
                        niz4[c]=niz4[j];
                        niz4[j]=temp;
                }
        }
}
                puts(niz2);
                printf("\n");
                puts(niz4);
                printf("\n&#37;d", br);

        scanf("%d", &sranje);
}

---

You forgot jJ in your consonant string...
and you've got cC in there twice...

QuantumPete

it's there now...:)

Ok, this is the same thing. You can see that code I have written up is working, but now i want to do it this way...

Code:

---

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

int main()
{
        int sranje;
        char niz[50];

        printf("Rijec: ");
        fgets(niz,sizeof(niz),stdin);

        funk(niz);
        scanf("%d", &sranje);
}

int funk(char niz[]){

        char *niz2, *niz3;
        int i,j=0,d=0;
        niz2=malloc(sizeof(char));
        niz3=malloc(sizeof(char));

        for(i=0;i<strlen(niz)+1;i++){

                if(niz[i]>=65 && niz[i]<= 122){

                        if(niz[i]=='A' || niz[i]=='E' || niz[i]=='I' || niz[i]=='O' || niz[i]=='U' ||
                                niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u'){
                               
                                niz2[j]=niz[i];
                                j++;}
                        else
                                niz3[d]=niz[i];
                                d++;
                                }
                        }
        printf("%s", niz2);
        printf("\n%s", niz3);
}

---

It's the same thing but little shorter. Everything works, well not really, if i input "Zadar", output is " aa""" ", instead of " is 2, and for niz3 is " z"d"r " and I want it to be aa, zdr. Any suggestions are most welcome.

You may want to do

Code:

---

niz2[j] = 0;
niz3[d] = 0;

---

before you try to print your strings.

I would probably do something like this:

Code:

---

    switch(tolower(niz[i]) {
      case 'a':
      case 'e':
      case 'i':
      case 'o':
      case 'u':
          niz2[..] = ....;
          break;
      default:
          niz3[...] = ...;
          break;
      }

---

--
Mats

Works fine for niz2[j], but niz3 is still messed up! What niz2[j]=0; does? And when I dont enter anything it still prints out some signs for niz3.

Quote:

---

Originally Posted by shardin

Works fine for niz2[j], but niz3 is still messed up! What niz2[j]=0; does? And when I dont enter anything it still prints out some signs for niz3.

---

Strings need to be terminated with a NULL character, otherwise the printf routine doesn't know where to stop outputting strings. It can't know after all, how much memory you've allocated. If you don't have a NULL terminator, then printf will continue reading memory past the end of your allocated memory until it encounters the first NULL char. That could be three chars from the end, it could be at the end of that page file or there could even not be any NULLs in memory.
So you should always set the last char in your string to NULL, by doing:

Code:

---

niz2[j] = 0;
/* or */
niz2[j] = '\0';

---

QuantumPete

Ok, can someone explain to me why this ...

Code:

---

for(i=0;i<strlen(niz)+1;i++){

                if(niz[i]>=65 && niz[i]<= 122){

                        if(niz[i]=='A' || niz[i]=='E' || niz[i]=='I' || niz[i]=='O' || niz[i]=='U' ||
                                niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u'){
                               
                                niz2[j]=niz[i];
                                j++;}
                        else
                                niz3[d]=niz[i];
                                d++;
                                }
                        }

---

...is not working, and this...

Code:

---

        for(i=0;i<strlen(niz)+1;i++){
                if(niz[i]>=65 && niz[i]<= 122){
                     
                                if(niz[i]=='A' || niz[i]=='E' || niz[i]=='I' || niz[i]=='O' || niz[i]=='U' ||
                                niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u')
             
                                niz2[j++]=niz[i];
                        else
                niz3[d++]=niz[i];
                                }
                        }

---

...works perfectly! In the upper code niz3 is messed up. Only difference is in...

Code:

---

{
                               
                                niz2[j]=niz[i];
                                j++;}
                        else
                                niz3[d]=niz[i];
                                d++;
                                }

---

...and...

Code:

---

                                niz2[j++]=niz[i];
                        else
                niz3[d++]=niz[i];

---

Only one of your strings is going to get a \0 written to the end of it.

Rearrange the code, and you get a different kind of luck.

> if(niz[i]>=65 && niz[i]<= 122)
How about
if(niz[i]>='A' && niz[i]<= 'z')

In this:

Code:

---

for(i=0;i<strlen(niz)+1;i++){

                if(niz[i]>=65 && niz[i]<= 122){

                        if(niz[i]=='A' || niz[i]=='E' || niz[i]=='I' || niz[i]=='O' || niz[i]=='U' ||
                                niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u'){
                               
                                niz2[j]=niz[i];
                                j++;}
                        else
                                niz3[d]=niz[i];
                                d++;
                                }
                        }

---

Check the braces - particularly on the else-branch. If you use Emacs, mark the region and do "M-x indent-region", or in MSVC under Edit->Advanced->Format Selection (again after marking the region).

Always using braces on ALL parts of if/else/while statements avoids this problem.

--
Mats

Quote:

---

> if(niz[i]>=65 && niz[i]<= 122)
How about
if(niz[i]>='A' && niz[i]<= 'z')

---

Or even

Code:

---

#include <ctype.h>

if(isalpha(niz[i]))

---

assuming that the intent of the code was to see if niz[i] was a lower or uppercase letter.

tolower() might also be called for -- then you don't have to check for 'E' and 'e', 'I' and 'i'.

> Always using braces on ALL parts of if/else/while statements avoids this problem.
Sound advice indeed!

Also:

Code:

---

for(i=0;i<strlen(niz)+1;i++){

---

Calculating strlen() is expensive, you might want to do it outside of the loop. And that's the same as

Code:

---

for(i=0;i<=strlen(niz);i++){

---

which might be easier to read, or it might not.

I'm not allowed to use string functions in exam! Only strlen!

No, the thing is in braces and 'else' statement! If I use niz[i]=niz2[a], i++, then I need to use braces, if I use niz[i++]=niz2[a], then I dont need braces and then affects 'else' statement, yes I just realized! It doesnt have anything with upper, lower...

Quote:

---

I'm not allowed to use string functions in exam! Only strlen!

---

I wasn't suggesting that you use a different function. I was suggesting that you calculate strlen() before the loop and save the result in a variable.

Code:

---

size_t len = strlen(niz);
for(i=0;i<len+1;i++){

---

Quote:

---

It doesnt have anything with upper, lower...

---

Your code also matches [, ], and other characters that are between 'Z' and 'a'. Is that what you wanted? Probably not. And your code isn't even portable. It could break with different character sets.

All in all, isalpha() is a very good idea.

The tolower() suggestion would make your code a lot easier.

Code:

---

                        if(niz[i]=='A' || niz[i]=='E' || niz[i]=='I' || niz[i]=='O' || niz[i]=='U' ||
                                niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u'){

---

->

Code:

---

                        char c = tolower(niz[i]);
                        if(niz[i]=='a' || niz[i]=='e' || niz[i]=='i' || niz[i]=='o' || niz[i]=='u'){

---

No, no, no....I think you are missing my point.

Quote:

---

No, the thing is in braces and 'else' statement! If I use niz[i]=niz2[a], i++, then I need to use braces, if I use niz[i++]=niz2[a], then I dont need braces and then affects 'else' statement, yes I just realized!

---

I realize that you identified the problem as being a case of missing curly braces, and that you can resolve this by using a single, more concise statement.

My suggestions are just that -- suggestions. They won't affect the compilation or execution (much) of your program, except to perhaps speed it up (in the strlen() case) or not count [ and ] etc as letters (in the isalpha() case). And the tolower() one makes the code shorter.

If you're not allowed to use functions from <ctype.h>, then . . . oh well. That would be ridiculous.

Quote:

---

If you're not allowed to use functions from <ctype.h>, then . . . oh well. That would be ridiculous.

---

Well, I'm not allowed. No, not on this exam. And no computer, we have to write it all on paper.

As I said, that's ridiculous. This is non-standard:

Code:

---

if(c >= 'a' && c <= 'z')

---

It's unportable because not all character sets have 'b' after 'a', etc. The only time you can use a range like that is between '0' and '9', which are guaranteed to be contiguous.

The portable way to do things is to use the functions in ctype.h.

For proof, all you have to do it look around the internet. Here's one source, Wikipedia: http://en.wikipedia.org/wiki/Ctype.h

Quote:

---

Early toolsmiths writing in C under Unix began developing idioms at a rapid rate to classify characters into different types. For example, in the ASCII character set, the following test identifies a letter:

Code:

---

if ('A' <= c && c <= 'Z' || 'a' <= c && c <= 'z')

---

However, this idiom does not work for other character sets such as EBCDIC.

---

I would seriously consider lodging a complaint to try to get them to let you use ctype.h stuff. There aren't any functions in there that can't be emulated with an if statement in an unportable, ASCII-dependent manner.

But it works fine with, "c>=65 && c<=122", those are letters from a -A to z-Z. That is acceptable.

No. Absolutely not. Did you read the article I linked to? It's completely unacceptable if you want your program to be portable. Sure, it works for ASCII. But it won't work for EBCDIC. It might not work for other character sets. It's simply not portable.

You can't assume on 'a'...'z' and 'A'...'Z' being contiguous.

Quote:

---

Originally Posted by dwks

No. Absolutely not. Did you read the article I linked to? It's completely unacceptable if you want your program to be portable. Sure, it works for ASCII. But it won't work for EBCDIC. It might not work for other character sets. It's simply not portable.

You can't assume on 'a'...'z' and 'A'...'Z' being contiguous.

---

And of course, we all have a few EBCDIC machines sitting around in our homes :-)

Are AS/400 machines using EBCDIC, or are they using ASCII/UNICODE?

--
Mats

Well I think the real point is not that we all might have some EBCDIC keyboards at home. It's more about not having to punch in some slightly mysterious-at-first-glance number code to talk about letters. Being able to refer to them reliably despite their order according to charmap.exe is simply a pleasant bonus that is always rewarded. Character constants make things easiest for us, the people.

Quote:

---

Originally Posted by citizen

Well I think the real point is not that we all might have some EBCDIC keyboards at home. It's more about not having to punch in some slightly mysterious-at-first-glance number code to talk about letters. Being able to refer to them reliably despite their order according to charmap.exe is simply a pleasant bonus that is always rewarded. Character constants make things easiest for us, the people.

---

Oh, sure, I agree that the above code should be (at least):

Code:

---

if (x >= 'A' && x < 'z')

---

Better yet:

Code:

---

if (isalpha(x))

---

as the latter doesn't include [\] and `.

But the fact is that 99.9% of all machines around today are using ASCII for the English parts of the characters - other letters outside that range isn't quite so easy.

The same applies to the constant "that won't work if the machine uses 1's complement for negative numbers". I've worked with computers for over 20 years and yet to see a machine that uses 1's complement. Yes, fine, there's machines that have been produced that uses different numerical formats, and if you wish to write code that is really portable, do so.

There's portable, and there is portable. Assuming that a machine runs with a ASCII (or closely enough related that the difference appears outside the English character set) character set is reasonable in all but very rare circumstances.

If you want portable code, the original code isn't internationalized either, so it wouldn't work in Swedish for example, since ÅÄÖ (and åäö of lower-case versions of course) are considered vowels in Swedish. Not sure if ü in German counts too. Certainly, French accented letters such as é and è or other decorated letters, e.g. ë would probably also count as vowels in their languages. So complaining about ONE part of a piece of code not being portable, when the rest of it is also utterly un-portable is pretty pointless.

--
Mats