CODEWITHSUNDEEP
cloopsif-statementnested-loops
EL02
EL02

Reputation: 322

Dynamic border of a rectangle

I want to create a hallow rectangle with a dynamic border, which means that if I input 3, then the border is going to be 3 columns thick. I've managed to do it with pre-given value, but I want the values to be inputted by the user.

#include <stdio.h>

int main()
{
    int i, j, n, m;
    printf("Rows: ");
    scanf("%d", &n);
    printf("Columns: ");
    scanf("%d", &m);
    
    for(i=1; i<=n; i++){
        for(j=1; j<=m; j++){
            if(i==1 || i==2 || i==3 || i==n-2 || i==n-1 || i==n || j==1 || j==2 || j== 3 || j==m-2 || j==m-1 || j==m){
                printf("*");
            }
            else{
                printf(" ");
            }
        }
        printf("\n");
    }

    return 0;
}

Upvotes: 0

Views: 43

Answers (1)

Beta
Beta

Reputation: 99144

Suppose you wanted to print a single row of 1000 spaces:

for(j=1; j<=1000; ++j){
  printf(" ");
}

So far, so good. Now suppose you want the first three characters to be asterisks:

for(j=1; j<=1000; ++j){
  if(j==1 || j==2 || j== 3){
    printf("*");
  }
  else{
    printf(" ");
  }
}

That works. There are other ways, but that works.

Now suppose you want the first 200 characters to be asterisks:

for(j=1; j<=1000; ++j){
  if(j==1 || j==2 || j== 3 || j==4 || ...

All right, this is clearly not the right approach. Good programmers dislike typing. Let's try one of the other ways:

for(j=1; j<=1000; ++j){
  if(j<=200)
    printf("*");
   else
    printf(" ");

That works. Here's another way, even better:

for(j=1; j<=200; ++j)
  printf("*");
for(j=201; j<=1000; ++j)
  printf(" ");

Now suppose you want the first 200 and last 200 characters to be asterisks:

for(j=1; j<=200; ++j)
  printf("*");
for(j=201; j<=1000-200; ++j)
  printf(" ");
for(j=1000-200+1; j<=1000; ++j)
  printf("*");

That should be enough to show you how to rewrite your code for rectangular borders.

Upvotes: 1

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