simplex minesweeper need help

[hergelenem]

i need a program which will generate a board(15x30) with randomly placed mines, and it will print the result on screen.

Example run:
>Enter N (number of mines): 20
>Here’s the random minefield:
> o o o o o o o o o o o o o o o o * * o o o o o o o o o o o o
> o o o o o o o o o o o * o o o o o o o o o o o o * o o o o o
> o o o * o o o o o o o o o o o o o o o o o o o o o o o o o o
> o o o * o o o o o o o o o o o o o o o o o o o o o o o o o o
> o o o o o o o o o o o o o o o * o o o o o o o o o o o o o o
> o o o o o o o o o o o o o o o o o o o o o o o o o o o * o o
> o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
> o o o o o o o o o o o * o o o o o o o o o o o o o o o o o o
> o o o o o o o o o o o * o o o o o o o o o o o o o o o o o o
> o o o o o o o o o o o o o o o o o o o o o o o o o o o * o o
> o o o o o o o o o o o o o o o o o o * o o o o o o o o o o o
> o o o o o o * o o o o o o o o o o o o * o o o o o o o o o o
> o o o o o o o o o o o o o * o o o o o o o o o o * o o o o o
> o o o o o o o o o o o o o o o o * o o o o o o o * o o o o o
> o o o o o o o o o o o o o o o o o o o o o * o o o * o o o o
-------------------------------------------------------------------------------
here is my code but it isn't perfect.it prints more than N value or less. can you help me?

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

int main (void)
{
int sayi,mayin,i,x,y,a,b;
int tarla[15][30];

printf ("Enter N (number of mines): ");
scanf ("%d", &mayin);



	tarla[a][b]=0;
for (sayi=0;sayi<mayin;sayi++)
{
srand(time(0));
do
{
x=1+rand()%16;
y=1+rand()%31;
}
while(tarla[x][y]==i);
tarla[x][y]=i;
}
for (x=0;x<15;x++)
{
for (y=0;y<30;y++)
{
if (tarla[x][y]==i) printf ("X");
else printf ("o");
printf("");
}
printf("\n");
}
return 0;
}

[Salem]

First, indent, so your code is readable.

#include <stdlib.h>

int main(void)
{
    int sayi, mayin, i, x, y, a, b;
    int tarla[15][30];

    printf("Enter N (number of mines): ");
    scanf("%d", &mayin);



    tarla[a][b] = 0;
    for (sayi = 0; sayi < mayin; sayi++) {
        srand(time(0));
        do {
            x = 1 + rand() % 16;
            y = 1 + rand() % 31;
        }
        while (tarla[x][y] == i);
        tarla[x][y] = i;
    }
    for (x = 0; x < 15; x++) {
        for (y = 0; y < 30; y++) {
            if (tarla[x][y] == i)
                printf("X");
            else
                printf("o");
            printf("");
        }
        printf("\n");
    }
    return 0;
}

See SourceForge.net: Indentation - cpwiki


Second, enable as many warnings as you can for your compiler,

$ gcc -W -Wall -ansi -pedantic -O2 foo.c
foo.c: In function ‘main’:
foo.c:8: warning: implicit declaration of function ‘printf’
foo.c:8: warning: incompatible implicit declaration of built-in function ‘printf’
foo.c:9: warning: implicit declaration of function ‘scanf’
foo.c:9: warning: incompatible implicit declaration of built-in function ‘scanf’
foo.c:16: warning: implicit declaration of function ‘time’
foo.c:31: warning: zero-length gnu_printf format string
foo.c:13: warning: ‘a’ is used uninitialized in this function
foo.c:13: warning: ‘b’ is used uninitialized in this function
foo.c:22: warning: ‘i’ may be used uninitialized in this function

All those "uninitialised" variables are bugs.

[Adak]

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

int main (void)
{
   int sayi,mayin,i,x,y;
                //int tarla[15][30];
   char tarla[15][30]; //ok

   printf ("Enter N (number of mines): ");
   scanf ("%d", &mayin);

   srand(time(0));  //ok


   for (sayi=0;sayi<mayin;sayi++)
   {
                           //srand(time(0));
                             //do
       x=1+rand()%16;
       y=1+rand()%31;
       tarla[x][y]='*';
                           //while(tarla[x][y]==i);
                            //tarla[x][y]=i;
   }
   for (x=0;x<15;x++)
   {
      for (y=0;y<30;y++)
      {
         if(tarla[x][y]!='*')
            tarla[x][y]= 'O');
         printf("%c",tarla[x][y];
      }
      printf("\n");
   }
   return 0;
}

I'd use a char array for this, and logic like the above. Make sense?

If you'll indent your code so subordinate lines of code are indented about 2-5 spaces, for each level of subordination, your code will be MUCH easier to study.

[hergelenem]

I'd use a char array for this, and logic like the above. Make sense?

If you'll indent your code so subordinate lines of code are indented about 2-5 spaces, for each level of subordination, your code will be MUCH easier to study.

normally i study that way,i just erased them to not cover much place

i rewrite code like this

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

int main (void)
{
   int sayi,mayin,i,x,y;
                
   char tarla[15][30]; 

   printf ("Enter N (number of mines): ");
   scanf ("%d", &mayin);



   for (sayi=0;sayi<mayin;sayi++)
   {
           srand(time(0))  ;             
    do{
       x=1+rand()%16;
       y=1+rand()%31;
       tarla[x][y]='*';
}while(tarla[x][y]==i);
     

 tarla[x][y]=i;

   }
   for (x=0;x<15;x++)
   {
      for (y=0;y<30;y++)
      {
         if(tarla[x][y]!='*')
         {  
         tarla[x][y]='O';
         printf("%c",tarla[x][y]);
      }
  }
      printf("\n");
   }
   return 0;
}

but it didn't placed any mines .

[Adak]

Look at this, and make your code very very similar:

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

int main (void)
{
   int sayi,mayin,i,x,y;
   char tarla[15][30]; //ok

   printf ("Enter N (number of mines): ");
   scanf ("%d", &mayin);
   (void) getchar();   //removes left over newline from keyboard buffer
   srand(time(0));  //ok

   for(x=0;x<15;x++) {
     for(y=0;y<30;y++) {
       tarla[x][y]='O';
     }
   }

   for (sayi=0;sayi<mayin;sayi++)
   {
       x=rand()%15;
       y=rand()%30;
       tarla[x][y]='*';
   }
   for (x=0;x<15;x++)
   {
      for (y=0;y<30;y++)
      {
         printf("%c",tarla[x][y]);
      }
      printf("\n");
   }
   
   (void) getchar(); //just holds the console window open - you may not need it
   return 0;
}

This is slightly simpler and runs correctly *except* you can wind up with cases where the random x and random y (because their range of numbers is so small), repeat to a square that already has a mine.

That will leave you one mine short, with every "collision" of mines.

How might you solve that problem?

[hergelenem]

This is slightly simpler and runs correctly *except* you can wind up with cases where the random x and random y (because their range of numbers is so small), repeat to a square that already has a mine.

That will leave you one mine short, with every "collision" of mines.

How might you solve that problem?

thank you very much it works fine