如何在C语言的控制台上打印叉号？

Question

I've just started studying information technologies and I am currently stuck on a programming assignment.

I have to write a code in C which displays a cross to the console, the size of the cross being determined by an initial input.

So the console output should look like this:

size?: 5(user input)

xooox

oxoxo

ooxoo

oxoxo

xooox

(replace the os with blank space)

I've now come as far as this:

#include <stdio.h>

int main(void)
{
    int n;

    printf("size?: ");
    scanf("%d",&n);

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if( (i==j) )
                printf("*");
            else 
                printf(" ");
        }
        printf("\n");
    }

    return 0;
}  

But this only displays one diagonal of the cross, I'm thinking that the opposite diagonal can be created by another condition after the if however I am lost as to what that condition might be.

Answer 1

You're definitely on the right track! Don't give up.

The way to think about this is to think about the loop counters. You've figured out one half of it. If the row and column are the same, you need to output a * . So what's the other condition? Well, think about counting backwards. If the row is the same as the column counted backwards, we also want a * .

I don't want to do your homework for you, so I'll hold off on writing the code, but hopefully that gives you a hint as to what you need to do.

Answer 2

Should you need some help, you have to check the column from the other side as well:

#include <stdio.h>
int n;

int main(void) {
    printf("size?: ");
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            if (i == j || i == n - j + 1) printf("*");
            else printf(" ");
        }
        printf("\n");
    }       
    return 0;
}  

Edit: There are suggestions to do i + j == n + 1 , but I believe i == n - j + 1 makes more sense, since:

• i is your current row
• j is your current column
• n is the size of your square (max row / max column)
• i == n - j + 1 means draw * at max - current + 1 column

Answer 3

Replace if( (i==j) ) with if( (i==j)||(i+j)==n+1 ) , That is,:

#include <stdio.h>

int main(void)
{
  int n;

  printf("size?: ");
  scanf("%d",&n);

  for(int i=1;i<=n;i++)
  {
    for(int j=1;j<=n;j++)
     {
      if( (i==j)||(i+j)==n+1 )
       printf("*");
      else 
       printf(" ");
     }
     printf("\n");
  }



   return 0;
 } 

Answer 4

First thing's first, C compilers don't like

for (int i = 0; i < n; i++)

They like

int i = 0;
for (i = 0; i < n; i++)

But if you're using a C++ compiler, you probably won't get that problem.

Now; back to solving the problem at hand!

On the line:

if( (i==j) )

With this conditional, you're plotting points at (1, 1), (2, 2), (3, 3) ...

You want to also plot points at (1, n), (2, n - 1), (3, n - 2) ...

So you need to add a second conditional to this if statement:

if ( (i==j) || (i == (n - j) + 1 ) )

Then you can simplify this up a bit if you want...

if ( (i==j) || (i == n - j + 1 ) )

And there you go! It now prints a cross, like you described in your question.