How to do this but in a bidimentional array

[endsamsara]

Hi, well i take lots of time figuring this out.

I need a program that tells me the coordinates of a word that is in other array,

what I have gotten is the y coordinate "the column", but
I need to do the case for a bidimentional array
get the x coordinate too "the row".

so this look like

hard //word to be found//

blablablablablabalblablabalbla //the puzzle where the word is
imtriyingtodomybestbutishard
blablablablablbalablabalabalbl

so the result i get from this is : (2,25) //row, column//
IN THE EXAMPLE I ONLY GET THE 25, AND I NEED TO GET THE 2 TOO

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


main()
{
 int coordenada(char *string, char *target);

 printf("%d\n", coordenada("mellevalachingadaporqueestoestamalplan", "mal"));
 getch();
 return 0;

}
 //returns the position of t in s//

 int coordenada(char *s,char *t)
 {
  char *p;

  for(p = s; *p != '\0';p++)    //p runs trough s //
      {
       char *x, *y;             //the first letter 'y' = 't'//

       x=p;                   //now keep going til find the end of the word//
       y=t;

    for(; *x != '\0' && *y != '\0' && *x == *y; x++, y++);

       if(*y == '\0')       //we get to the end of the word

          return p - s + 1;    //this gives the coordinate 'y'//


      }
      return -1;
             //if it fails//
 }

PLEASE HELP ME , I NEED TO FINISH THIS BY THE END OF THIS DAY
PLEASE!!:-|

[Narue]

You're not even using a two dimensional array. But if you were, did it occur to you that you can do this?

for ( i = 0; i < ROWS; i++ ) {
  int col =  coordenada ( array2d[i], target );

  if ( col != -1 )
    printf ( "\"%s\" found at (%d,%d)\n", target, i, col );
}

[endsamsara]

thanks, sorry for so many questions, but im newbie

how do i define the array2d

how could i use the code you have in my code??? does it go in first place?? dont know how to put all in place.

like:

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


main()
{
 int coordenada(char *string, char *target);

 printf("%d\n", coordenada("mellevalachingadaporqueestoestamalplan", "mal"));
 getch();
 return 0;

}
 //returns the position of t in s//

 int coordenada(char *s,char *t)
 {
  char *p;
   
  //////?????????????///
  int i;
  char ROWS;
  char col;
  char array2d[3][10]
for ( i = 0; i < ROWS; i++ ) {
  int col =  coordenada ( array2d[i], target );

  if ( col != -1 )
    printf ( "\"%s\" found at (%d,%d)\n", target, i, col );
}

/////////////'?//

  for(p = s; *p != '\0';p++)    //p runs trough s //
      {
       char *x, *y;             //the first letter 'y' = 't'//

       x=p;                   //now keep going til find the end of the word//
       y=t;

    for(; *x != '\0' && *y != '\0' && *x == *y; x++, y++);

       if(*y == '\0')       //we get to the end of the word

          return p - s + 1;    //this gives the coordinate 'y'//


      }
      return -1;
             //if it fails//
 }

[iamthwee]

If you are only just realising you have to use a 2d array now, you've no hope of getting this done by today.

:lol:

[endsamsara]

well i dont know exactly how to define it well

[Narue]

>but im newbie
I don't care. You ask questions, I answer them. If I need to know your level of experience, I'll ask a pointed question to determine that. Calling yourself a newbie serves no purpose at all.

>how could i use the code you have in my code???
I was thinking that you define the array and just paste the code I gave you into main as well:

#include <stdio.h>

#define ROWS 3

int coordenada ( const char *s, const char *t )
{
  const char *p;

  for ( p = s; *p != '\0'; p++ ) {
    const char *x = p, *y = t;

    for ( ; *x != '\0' && *y != '\0' && *x == *y; x++, y++ )
      ;

    if(*y == '\0')
      return p - s + 1;
  }

  return -1;
}

int main ( void )
{
  const char *array2d[] = {
    "blablablablablabalblablabalbla",
    "imtriyingtodomybestbutishard",
    "blablahardlablbalablabalabalbl"
  };
  const char *target = "hard";
  int i;

  for ( i = 0; i < ROWS; i++ ) {
    int col =  coordenada ( array2d[i], target );

    if ( col != -1 )
      printf ( "\"%s\" found at (%d,%d)\n", target, i, col );
  }

  return 0;
}

It's not exactly brain surgery if you have a book on C.

[endsamsara]

thank you very much

[endsamsara]

Hello, huehuehe took some time to solve this, like when you know how to program is easier, so, thanks to all.

here es the code.

Hola, me puse a resolver esto, ya que cuando iniciaba, no tenía ni idea de como hacerlo, ahora me contesto a mi mismo.
este es el codigo:

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
#include <string.h>
#include <ctype.h>
#define MAXLINEA 180 //max tamaño de una linea

//busca una palabra en una linea, si esta, regresa la coordenada 'x'
int coordenada ( const char *linea, const char *palabra ){
    const char *p, *x,*y;
    for ( p = linea; *p != '\0'; p++ ) {
        x = p;
        y = palabra;
        for ( ; *x != '\0' && *y != '\0' && *x == *y; x++, y++ )
            ;
            if(*y == '\0')
                  return p - linea + 1;
    }
return -1;
}

//reservamos espacio para nuestro arreglo 'bidimensional'
char **reservaEspacio(int filas, int cols){
   char **arreglo;
   int i;
   /* espacio para el arreglo de apuntadores a entero*/
   arreglo = (char**)malloc(filas * sizeof(char*));
   if(arreglo == NULL){
      fprintf(stderr, "Error 1-d \n");
      exit(-1);
   }                                                                               
   /* espacio para los arreglos de letras */
   for(i = 0; i < filas; i++){
      arreglo[i] = (char*)malloc(cols * sizeof(char));
      if(arreglo[i] == NULL){
         fprintf(stderr, "Error 2-d\n");
         exit(-1);
      }
   }                                                                               
   return arreglo;
}

//regresa un arreglo 'bidimensional', llenandolo a partir
//de un archivo de entrada 'entrada.txt'
char **lee (int *numPalabrasPtr, char *nombreArchivo) {
  char linea[MAXLINEA];   
  char **arreglo;
  int i;
  FILE *f;
  if ( ! ( f = fopen (nombreArchivo, "r") ) ) {
    printf ("error al abrir archivo %s\n", nombreArchivo);
    exit (-1);
  }
   while ( fgets (linea, MAXLINEA, f) ) {
         *numPalabrasPtr = *numPalabrasPtr+1;
  }
  arreglo = reservaEspacio (*numPalabrasPtr, MAXLINEA);
  rewind (f);
  for ( i=0 ; i <*numPalabrasPtr ; i++ ) {
    fgets (arreglo[i], MAXLINEA, f);
    arreglo[i][strlen(arreglo[i])] = '\0';
  }
  fclose (f);
  return (arreglo);
}

//busca la palabara en el arreglo, si esta,la imprime
// y escribe en un archivo 'salida.txt'
void busca(int renglones, char **array2d , char *target){
     FILE *sal;
     sal = fopen("salida.txt","a+");     
     int i, col;
     for ( i = 0; i < renglones; i++ ) {
         col =  coordenada ( array2d[i], target );
         if( col != -1 ){
             fprintf(sal,"\"%s\" \t encontrada en (%d,%d)\n",target, i+1, col);             
             printf ( "\"%s\" \t encontrada en (%d,%d)\n", target, i+1, col );            
         }    
     }
      fclose(sal); 
}


//inicializamos todo, y decimos de que archivo se van a buscar las palabras
int main (){
  int numPalabrasPtr = 0;
  int i=0;
  char **array2d;
  //numPalabrasPtr se pasa por referencia, para saber cuantas lineas tiene el archivo
  array2d  = lee(&numPalabrasPtr, "entrada.txt"); 
  char *palabrasBusca[] = {"gato", "perro", "raton", "elefante", "rino", "serpiente", "pescadito"};
  //de q tamaño es la lista??
  int len = sizeof(palabrasBusca) / sizeof(char *) ;
  //buscamos cada palabra en nuestra matriz de palabras
  for(i =0; i< len; i++){
        busca(numPalabrasPtr,array2d, palabrasBusca[i] );
  }
  getch();
  return 0;
}

el archivo entrada usado fue:

aaaaaaaaaaaaaaaaaaaaa
setrsdfdsrrtdpyrinoeq
weraderelefantewwerrr
trtevhjujhaspescadito
rtxvfdghhgperrodrdvbh
ifghhfgaaasdserpiente
naendsdsadsasafrrsdft
nssdofgfgghghghdddddd
ttegatovvvfgfyhgggggg
rrrrrrrrrrrrrrrrrrrrr