How to remove characters from console output in C

Question

Code1:

#include<stdio.h>
int main()
{
    unsigned short i, j, k;
    for (i = 0; i < 2; i++)
    {
        k = i * 4 + 4;

        for (j = k - 4; j < k; j++)
            printf("%hu ", j);

        putchar('\n');
    }

    return 0;
}

Output of Code1:

0 1 2 3 
4 5 6 7 

Remarks of Code1:

1. Space after 3 and 7
2. Newline after 7 (Stackoverflow has removed it)

Code2:

#include<stdio.h>
int main()
{
    unsigned short i, j, k;
    for (i = 0; i < 2; i++)
    {
        k = i * 4 + 4;
        for (j = k - 4; j < k; j++)
        {
            printf("%hu", j);

            if (j + 1 != k) putchar(' ');
        }

        if (i + 1 != 2) putchar('\n');
    }

    return 0;
}

Output of Code2:

0 1 2 3
4 5 6 7

Remarks of Code2:

1. No space after 3 and 7
2. No newline after 7

Additional remark of Code2:

The problem of Code2 is that the algorithm always compares two values in the if blocks and so Code2 is not efficient. I want to use Code1 and change these:

Code3

#include<stdio.h>
int main()
{
    unsigned short i, j, k;
    for (i = 0; i < 2; i++)
    {
        k = i * 4 + 4;

        for (j = k - 4; j < k; j++)
            printf("%hu ", j);

        printf("\b\n");
    }

    putchar('\b');
    return 0;
}

These do not show the same output of Code2 because \b does not erase anything.

My question:

Is there any standard way to do so what I've tried in Code3?

Remark of my question:

1. I have searched the internet but have not determined the solution.
2. Edit the question if it is not clear.

Edit: I don't know why my question is not useful or constructive. Though the above is an arbitrary small example, but performance might be an issue when processing very large amount of data. I thought that the way of removing character from console output might improve performance and there might be a specific way to do so. That's why I've asked the question. I could write the following codes in the answers. Now I've known via comments that removing character from console output is not possible because it is implementation dependent.

Answer 1

The usual approach to this is to treat either the first or the last printf as a special case (outside of the loop):

for(ii=0; ii<2; ii++) {
  jj = 0;
  printf("%d", jj);      // first number printed without space.
  for(jj=1; jj<4; jj++) {
    printf(" %d", jj);   // include the space before the number printed
  }
  if(ii<2-1) printf("\n");
}

Obviously I simplified how the loops are constructed and what is printed - for simplicity. You could make the first printf statement

printf("\n%d", jj);

then you have a newline at the start of your output (often a good thing) and then you don't need the if statement later - you just don't have a newline printed at the end of the line (because it will be printed at the start...)

There are marginally more efficient ways of doing this that would involve no if statements at all - but these all come at the expense of less readable code. For example, here is a "no loop unrolling and no additional if statements" version of the code:

http://codepad.org/01qPPtee

#include <stdio.h>
int main(void) {
int ii, jj;
   ii = 0;
   while(1) {
      jj = 0;
      while(1) {
        printf("%d", jj);   // include the space before the number printed
        jj++;
        if(jj<4) printf("."); else break;
      }
      ii++;
      if(ii<2) printf("*\n"); else break;
    }
return 0;
}

Output:

0.1.2.3*
0.1.2.3

Basically I have taken the functionality of the for loop and made it explicit; I also use a . rather than a and "*\n" rather than "\n" to show in the printout that things behave as expected.

It does what you asked without extra evaluation of the condition. Is it more readable? Not really...

Answer 2

If it really bothers you, you can unroll your loops a little so that you treat the last item as a special case:

#include<stdio.h>
int main()
{
    unsigned short i, j, k;
    for (i = 0; i < 1; i++)
    {
        k = i * 4 + 4;
        for (j = k - 4; j < k - 1; j++)
        {
            printf("%hu ", j);
        }
        printf("%hu\n", j);
    }

    k = i * 4 + 4;
    for (j = k - 4; j < k - 1; j++)
    {
        printf("%hu ", j);
    }
    printf("%hu", j);

    return 0;
}

Answer 3

#include <stdio.h>

int main(){
    unsigned short num = 0, to=8;

    while(num < to){
        printf("%hu", num++);
        if(num < to)
            putchar(num % 4 == 0 ? '\n' : ' ');
    }
#if 0
    do{
        printf("%hu", num++);
    }while(num < to && putchar(num % 4 == 0 ? '\n' : ' '));
#endif
    return 0 ;
}

Answer 4

Well, to try to answer your question, here's how I would do it:

for (i = k = 0; i < 2; i++){
  if (i > 0) printf("\n");
  for (j = 0; j < 4; j++, k++){
    if (j > 0) printf(" ");
    printf("%d", k);
  }
}

I do it this way because I want to be sure every line but the first starts with a \n, and every item is separated by a space from the one before it. Also, I do not want the row and column position to be intimately tied to the content of what is being printed.

In terms of performance, keep in mind that these if statements cost about 1 cycle, while each character printed costs at least hundreds if not thousands. printf goes through many layers of system calls to interpret its format string, build a buffer, send the buffer to the system I/O routines, which then cause repainting and scrolling of the console window. Get the idea?

DO NOT WORRY about performance unless you know you have a problem. Then, don't guess. Use a diagnostic. Here's what I do.