Finding character in string

Question

Given a string and a character, I have to find how many times the character exists in the string.

This is what I have so far:

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

main ()
{
  char string[10];
  char c1, c2;
  int contador, i, l, n;

  printf ("Introduza uma string e dois caracteres.\n");
  scanf ("%s %c %c", &string[i], &c1, &c2);

  l = strlen (string);
  contador = 0;

  for (n = 0; n < l; n++)
    {
      if (c1 == string[i])
    {
      contador = contador + 1;
    }
    }
  printf ("%d\n", contador);
}

The text in printf is in portuguese, and it means "Introduce a string and two characters". The second character is there for later.

Would appreciate any help you can give.

Okay, so what's wrong with the code you have now? Is it not working? What is it doing? — tckmn
– tckmn, Commented Nov 3, 2013 at 13:48
possible duplicate of Replacing characters in strings malloc — Alexej Magura
– Alexej Magura, Commented Dec 16, 2014 at 4:41

haccks · Accepted Answer · 2013-11-04 13:27:15Z

3

You have to made some changes:
Change

 scanf ("%s %c %c", &string[i], &c1, &c2);

to

 scanf ("%s %c %c", string, &c1, &c2);

and

 if (c1 == string[i])

to

 if (c1 == string[n])

Also you can keep if (c1 == string[i]) unchanged by changing for (n = 0; n < l; n++) to for (i = 0; i < l; i++).
Here is the modified code snippet

printf ("Introduza uma string e dois caracteres.\n");
scanf ("%s %c %c", string, &c1, &c2);  


l = strlen (string);
contador = 0;

for (n = 0; n < l; n++)
{
      if (c1 == string[n])
      {
          contador = contador + 1;
      }
}

edited Nov 4, 2013 at 13:27

answered Nov 3, 2013 at 13:52

haccks

107k28 gold badges181 silver badges274 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

cyphar · Accepted Answer · 2013-11-03 13:53:37Z

1

Here is a simple implementation of a function that does what you need.

int strnchr(char *string, char ch) {
    int i, len = strlen(string), found = 0;
    for(i = 0; i < len; i++)
        if(string[i] == ch)
            found++;
    return found;
}

answered Nov 3, 2013 at 13:53

cyphar

2,9301 gold badge17 silver badges25 bronze badges

Comments

Community · Accepted Answer · 2017-05-23 10:24:58Z

0

Here is the modified code, and the 'why' is haccks's answer.

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

int main ()
{
  char string[10];
  char c1;
  int contador, l, n;

  printf ("Introduza uma string e dois caracteres.\n");
  scanf ("%s %c", string, &c1);

  l = strlen (string);
  contador = 0;

  for (n = 0; n < l; n++)
    {
      if (c1 == string[n])
          contador++;
    }

  printf ("%d\n", contador);

  return 0;
}

edited May 23, 2017 at 10:24

CommunityBot

11 silver badge

answered Nov 3, 2013 at 15:18

keyring

1

Comments

jocke-l · Accepted Answer · 2013-11-03 15:24:16Z

0

If you need a fast implementation of this function you can use the strchr-function from the standard c library, which should be well optimised on most common systems.

int strnchr(char *str, char ch) {
   int i;

   for (i = 0; (str = strchr(str, ch)) != NULL; i++) {
      if (++str == '\0') {
         break;
      }
   }

   return i;
}

answered Nov 3, 2013 at 15:24

jocke-l

7333 silver badges8 bronze badges

Collectives™ on Stack Overflow

Finding character in string

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related