Thread: function and strings - C

Thanks for your answers
@zacs
this is not homework. I found the exercise(as many others) in a site and i am trying to do it for practise only

i will try your tips and if i have a problem i will post again.So see you soon in a few hours

Originally Posted by zacs7

Since this is homework (probably) you're not getting the answer, but here is a pretty good start:

Code:

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

static void printUnique(const char * a, size_t m, const char * b, size_t n);

int main(void)
{
    char str1[] = "Hello";
    char str2[] = "Hallo";

    /* what's in str1 that isn't in str2 ? */
    printUnique(str1, strlen(str1), str2, strlen(str2));

    /* what's in str2 that isn't in str1 ? */
    printUnique(str2, strlen(str2), str1, strlen(str1));

    return 0;
}

static void printUnique(const char * a, size_t m, const char * b, size_t n)
{
    size_t i, f;
    int found = 0;

    /* print unique from a that aren't in b */
    for(i = 0; i < m; i++)
    {
        found = 0;
        for(f = 0; f < n; f++)
        {
            if(a[i] == b[f])
            {
                found = 1;
                break;            /* no point checking the rest */
            }
         }
        if(found == 0)
            printf("%c is in a but not b...\n", a[i]);
    }
    return;
}

/* output:
e is in a but not b...
*/

If you have to use pointers just use a pointer to point at a[i], ie

Code:

char * ptr;

/* ... */
for(i = 0; i < m; i++)
{
    ptr = a[i];
    printf("value is %c...\n", *ptr);
/* ... */

Yes i think this way will work too, its a kind of brute force attack and it is great. But why not to use strpbrk whick imho does the same job. Well its your choice anyway.

i made another try.Still i don't use malloc but an array.If this code work after i write it with malloc.The problem is that it doesn't print me nothing

Code:

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

char ar[20];

void fun(char *s1, char *s2);

int main(void)
{
 char *p1="hellosw";
 char *p2="ahellogw";

 fun(p1,p2);

 printf(ar);
 return 0;
}

void fun(char *s1, char *s2)
{
 int i,j,l;
 int found;
 int slen1,slen2;

 slen1=strlen(s1);
 slen2=strlen(s2);

 found=0; 
 l=0;

 for(i=0; i<slen1; i++)
 { for(j=0; j<slen2; j++)
   { if(s1[i]==s2[j]) /*for each element of s1[i] i check if there is also in s2*/
       found=1;
   }
  if (found=0)  
    ar[l++]=s1[i]; /*if not i copy the character to array ar*/
    
}

}

where is the mistake?

use

Code:

if (found==0)

or just

Code:

if (!found)

Kurt

still doesn't print anything with found==0

Code:

for(i=0; i<slen1; i++)
 { 
   found = 0;
   for(j=0; j<slen2; j++)
   { if(s1[i]==s2[j]) 
      { found=1; break;}
       
   }
  if (found==0)
      ar[l++]=s1[i]; 
}

ssharish

yes now prints me the hole first string and i don't want that
i think that in the loops i am making these:
if s1="het" and s2="aht"
for i=0
for j=0
s1[0]=s2[0] (h=a) (false)
for j=1
s1[0]=s2[1] (h=h) (true and break)

for i=1
for j=0
s1[1]=s2[0] (e=a) (false)
for j=1
s1[1]=s2[1] (e=h) (false)
for j=2
s1[1]=s2[2] (e=t) (false)
then
found is still 0
I am putting e in ar[0]

etc.

Does my code do these things that i am describing to you or not?
If yes then why don' t print what i want?

i think you are looking for this

Code:

 for(i=0; i<slen2; i++)
 { 
   found = 0;
   for(j=0; j<slen1; j++)
   { if(s1[j]==s2[i]) /*for each element of s1[i] i check if there is also in s2*/
      { found=1; break;}
       
   }
  if (found==0)
  {  
    ar[l++]=s2[i]; /*if not i copy the character to array ar*/
  }

ssharish

You have to reset found to 0 inside the outher loop.
Kurt

Code:

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

char ar[20];

void fun(char *s1, char *s2);

int main(void)
{
 char *p1="hello";
 char *p2="hallo";

 fun(p1,p2);

 printf(ar);
 getchar();
 return 0;
}

void fun(char *s1, char *s2)
{
 int i,j,l;
 int found;
 int slen1,slen2;

 slen1=strlen(s1);
 slen2=strlen(s2);

 found=0; 
 l=0;

 for(i=0; i<slen1; i++)
 { 
   found = 0;
   for(j=0; j<slen2; j++)
   { if(s1[i]==s2[j]) /*for each element of s1[i] i check if there is also in s2*/
      { found=1; break;}
       
   }
  if (found==0)
  {  
    ar[l++]=s1[i]; /*if not i copy the character to array ar*/
  }  
}

}

/* my output
e
*/

You get the output. What else u need?

ssharish

YES why do you use getchar??
In my code i don't use getchar.
Can you explain me?
With getchar the code runs as i want but why?

Because i need to Stop my Windows Console from disappearing everytime I run my program. Since I am using Dev-C++.

You could ignore that, if you want. If you are on console.

ssharish

Thanks i found the problem .It works also without getchar