Star/character Box

[rmathus]

I need a little help with this program. I need it to ask the user to enter a number and then a character. It will print out for example with values 5 and R,
RRRRR
R___R
R___R
R___R
RRRRR

(where the _ is white space)

I know I need to use another while or for loop (I'm guessing while), but I can't figure out where to put it. Here's the portion of the code that needs the work.
Thanks guys

int printSquare(int size, char character)
{
int i = 1;
int counter = 1;  
   for (i = 0; i < size; i++)
   {
      while (counter <= size)
      {
         counter++;
         printf("%c", character);
      }

   printf("\n");  
   counter = 1;
   }  
return;
}

It currently prints out:
RRRRR
RRRRR
RRRRR
RRRRR
RRRRR.

Any hints or suggestions? Thanks again!

[Prelude]

Don't try to use just a single nested loop to do it all, instead use several loops. One to print the top line, one to print the bottom line, and one with another nested inside to handle the lines with spaces in the middle:

#include <stdio.h>

void printsquare ( int width, int height, char c )
{
  int i;
  int j;

  for ( i = 0; i < width; i++ )
    putchar ( c );
  putchar ( '\n' );
  for ( i = 1; i < height; i++ ) {
    putchar ( c );
    for ( j = 1; j < width - 1; j++ )
      putchar ( ' ' );
    printf ( "%c\n", c );
  }
  for ( i = 0; i < width; i++ )
    putchar ( c );
  putchar ( '\n' );
}

int main ( void )
{
  printsquare ( 5, 5, '*' );

  return 0;
}

[Jeremy G]

Actaully, a Single nested loop would work fine:

void drawSquare(int s, char c) {
	for( int i = 0; i < s; i++ ) {
		for( int j = 0; j < s; j++ ) {
			if( (i == 0 || i == s-1) || (j == 0 || j == s-1) )
				printf("%c",c);
			else
				printf(" ");
		}
		printf("\n");
	}
}
// use me like this in main program:
drawSquare( 5, 'R');

:P

P.S: FIgured I'd put SOME explination to the code
basicly the nested loop gives you the sqaure traversal row by column.

i == 0
// this accounts for any and all cases where we are in the top row of the square
i == s-1
// this accounts for any and all cases where we are in the bottom row of the sqaure
j == 0
// this accounts for any and all cases where ew are in the left most column of the square
j == s-1
// this accounts for any and all cases where we are in the right most column of the square
// any other situation we are inside the actualy sqaure boundry, so we draw a space.

[Prelude]

>Actaully, a Single nested loop would work fine
I didn't say it wouldn't work, I meant that a single nested loop would be more difficult to visualize. There are a number of ways to get the correct output.

[Jeremy G]

Originally posted by Prelude
>Actaully, a Single nested loop would work fine
I didn't say it wouldn't work, I meant that a single nested loop would be more difficult to visualize. There are a number of ways to get the correct output.

All I ment was that it isn't that difficult to come up with a single nested loop solution. I merely think its a better practice to go a different rout then the easiest way. Im not makin this a personal attack on you or your code. I think it woulda been better to give a solution more related to what he was trying to do then to give him and entirely different solution. This way he can find/realize the error in his solution, and fix it.

[rmathus]

Awesome! Sorry it took so long to reply. I appreciate the help from the both of you, Prelude and dbgt goten! Thanks again!