"How many word of the same kind in a sentence " program ..?

Hello ,
This is a program intends to tell the number of the each word from the sentence. For example If write

naber aga naber naber aga

output will be like this:

naber 3
aga 2

I have written a program , I have a little problem I take a output like this :
naber 3
aga 2
naber 3
naber 3
aga 2

I know why I take this problem I have a solution . This is the codes by the way :

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void sonuc(char *,char *,int,int *);
int main() {
	char *s=malloc(40*sizeof(char));
gets(s);
       char *b,c[40];
	char *p[10];	
	char a[40];
	strcpy(a,s);
	b=strtok(a," ");
	int y=0;
	int x;
strcpy(p[0],b);
	printf("%s",p[0]);
	int i=1;
	int m=0;
	while(b!=NULL) {
       x=strlen(b);
for(m=0;m<i;m++) {
			if(strcmp(p[m],b)==0) {
                              b=strtok(NULL," ");
				continue; }
		}
			strcpy(c,s);
		sonuc(c,b,x,&y);
                 printf("%s kelimesinden %d tane var\n",b,y);
		y=0;
            b=strtok(NULL," ");
		i++;
		}
	return 0;
}





		void sonuc(char *d,char *str,int t,int *ptr) {
if(strstr(d,str)==NULL) return;
else {
  d=strstr(d,str);
			*ptr+=1;
			sonuc(&d[t],str,t,ptr); }
}

algorithm is not really hard. I am sure you will figure it out easily.

I have a solution for the problem . I can save each word into a struct or pointer string then I can compare the current word with the former words I have passed.If there is a same I can stop the loop to the beginning.

I want to know if there is another solution or algorith for tis program which is far more simpler than mine.

is my language hard to understand?

I want to know if there is another solution or algorith for tis program which is far more simpler than mine.

yes, there is.

but that's the point of learning how to develop software, is figuring out better ways. you want to be told the part we expect you to enjoy the most?! hah! never...

... well... i'm not taking the time to re-write your software to show how. up to you to discover the solutions. enjoy learning c++!

actually, maybe you want to learn math/algorithms, that would be more ideal of a tool for this sort of question, as far as breaking things down intuitively go anyway.

check the return value of malloc.

http://faq.cprogramming.com/cgi-bin/...&id=1043284351

I'm quite amazed to see how some people are able to write such hard-to-read (and understand) code. I bet you had some compilation error/warning and semantic bug the first time you compiled/ran the source code. If not, duh, you are a real genious (truly). Those variable name makes so little sense, and what about the indentation! This is truly amazing, i don't think i could do something like this, men, it just look too complex.

By the way this ain't C. And there's some "don't do this" in your code, like this

Code:

char *p[10];
strcpy(p[0],b); // uh oh

Anyway

I am not genious for sure , it is just the way I do things . Ununderstandable ways. Anyway what is with that strcpy thing? Ofcourse you cant do p[0]? =) I really didnt get the point.

I am sorry or variables are turkish ,
%d kac tane var means
"Number of ..... is %d".

Originally Posted by ozumsafa

I am not genious for sure , it is just the way I do things . Ununderstandable ways. Anyway what is with that strcpy thing? Ofcourse you cant do p[0]? =) I really didnt get the point.

What is p[0] pointing at?

--
Mats

yea I am talking about that I didnt do such thing , I did not undertand why he mentioned that.

He means that you can't just use char *p[10] like it's some magical place to shove text. Not only does the array have to have enough pointers but they should always point to some memory in order to be useful.

Maybe an example will help.

Code:

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

char *strdupl( const char orig[] ) {
  if ( orig ) {
    char *dup = malloc( strlen( orig ) + 1 );
    strcpy( dup, orig );
    return dup;
  }
  else {
    return NULL;
  }
}

int main( ) {
  size_t i, count;
  const char parseme[] = "C:/Documents and Settings/Owner/My Documents/";
  const char tok[] = ":/ ";
  char *parsable = strdupl( parseme );
  char *parts[8];
  char *seg;

  count = 0;
  printf( "Breaking apart \"%s\" by \"%s\":\n\n", parseme, tok );
  for ( seg = strtok( parsable, tok ); seg != NULL; seg = strtok( NULL, tok ) ) {
    parts[count] = seg;
    count++;
  }
  parts[count] = NULL;

  for ( i = 0; i < count; i++ ) {
    puts( parts[i] );
  }
  free( parsable );
  return 0;
}

Breaking apart "C:/Documents and Settings/Owner/My Documents/" by ":/ ":

C
Documents
and
Settings
Owner
My
Documents

Originally Posted by ozumsafa

yea I am talking about that I didnt do such thing , I did not undertand why he mentioned that.

It looks like this in the beginning of the code you posted:

Code:

gets(s);
       char *b,c[40];
	char *p[10];	
	char a[40];
	strcpy(a,s);
	b=strtok(a," ");
	int y=0;
	int x;
strcpy(p[0],b);
	printf("%s",p[0]);

[I added the red]

--
Mats