Remove white space at the end of the output in C

Question

The following code is for printing the elements of a matrix in spiral order. The program works fine. The problem, however, is that the online compiler against which I\'m checkin

Answer 1

An answer after accepted answer:

Rather than:

while (k < m && l < n) {
  ...
  printf("%d ", a[k][i]);
  ...
  printf("%d ", a[i][n-1]);
  ...
  printf("%d ", a[m-1][i]);
  ...
  printf("%d ", a[i][l]);
  ...
}

Change the format.

const char *format = "%d";
while (k < m && l < n) {
  ...
  printf(format, a[k][i]);
  format = " %d";
  ...
  printf(format, a[i][n-1]);
  ...
  printf(format, a[m-1][i]);
  ...
  printf(format, a[i][l]);
  ...
}
fputc('\n', stdout); // if desired.

Answer 2

You can set a boolean variable isFirst to true before your printing out any stuff, and test it before each printf statement. If isFirst, do not print a space but set isFirst to false; else print a single space. After that, continue with printing your number without a space.

Alternative: Instead of printing your results immediately, create a results array. Store your results in there, and when done, print out the results in a tight loop. You can print the first number without a leading space, then loop over the remainder and print them with a leading space.

Answer 3

Looking at your code, this for (the first one in the while loop):

for (i = l; i < n; ++i)
    printf("%d ", a[k][i]);

will always be executed at least ones (because l<n, coming from the while's condition).

Then you can just do the following:

• always add the space in front of the number
• add a single if check just for this very first for-loop (use some bool flag).

For example, something like:

bool first = true;
while (k < m && l < n)
{
    for (i = l; i < n; ++i)
    {
        if( ! first )
        {
            printf(" %d", a[k][i]);
        }
        else
        {
            printf("%d", a[k][i]);
            first = false;
        }
    }
    // ....
}

This will be rather efficient and short solution - the if is in just one loop and the flag will be true just once (will avoid cache misses).

Answer 4

You can put an if condition in your for loops

for (i = l; i < n; ++i)
{
        printf("%d", a[k][i]);
        if(i < n-1)
            printf(" ");
}

Answer 5

the printf() statement,

 printf("%d ", a[k][i]);

results in extra space. use

"%d" without space or use space in the begining as,

" %d"

then at the end there wont be a extra space.
its about how you use space in your printf(). use space in a way that extra space is not present at the end as you wanted.
You can use code like this,

while (k < m && l < n)
    {
        for (i = l; i < n; ++i)
        {
            if(l==0&&i==0)
            {
              printf("%d", a[k][i]);
            }
            else
              printf(" %d", a[k][i]);
        }
        k++;
        for (i = k; i < m; ++i)
            printf(" %d", a[i][n-1]);
        n--;
        if ( k < m)
        {
            for (i = n-1; i >= l; --i)
                printf(" %d", a[m-1][i]);
            m--;
        }
        if (l < n)
        {
            for (i = m-1; i >= k; --i)
                printf(" %d", a[i][l]);
            l++;
        }

Answer 6

You need to suppress the space when you don't need one. You can do it like this:

Add these declarations:

char *format = "%d";
int first_number = 1;

Add this after the first printf:

if (first_number) {
    /* Now we want a space between numbers */
    first_number = 0;
    format = " %d";
}

Change your printf:s to use the new variable:

            printf(format, ...);