understanding detab in K&R

Question

I'm teaching myself C and working through K&R. I'm doing exercise 1-20:

Write a Program entab that replaces strings of blanks by the minimum number of tabs and blanks to acheive the same spacing. Use the same tab stops as for detab.

I worked through this program myself, but was reviewing other solutions:

#include<stdio.h>
#define TABINC 8

int main(void)
{
  int nb,pos,c;

  nb = 0;
  pos = 1;

  while((c=getchar())!=EOF)
  {
    if( c == '\t')
    {
      nb = TABINC - (( pos - 1) % TABINC);   // <---- How does this work

      while( nb > 0)
      {
        putchar('#');
        ++pos;
        --nb;
      }
    }
    else if( c == '\n')
    {
      putchar(c);
      pos = 1;
    }
    else
    {
      putchar(c);
      ++pos;
    }
  }

  return 0;
}

I have difficulty understanding how this part works nb = TABINC - (( pos - 1) % TABINC);. Can someone please explain what this is exactly doing step by step? Perhaps walk me through an example?

Thank you.

Answer 1

Consider the following input text:

\tone\ntwo\tthree\nsixteen\tseventeen\teighteen\n

This is what we want it to produce (tabs replaced with ···):

         Column: |       |       |       |
         123456789012345678901234567890123
Line: 1  ········one
      2  two·····three
      3  sixteen·seventeen·······eighteen

The number of spaces required for each tab character is always going to be a number from 1 to TABINC (i.e., 8) inclusive. The tabs in the above example are expanded as follows:

Current   Next tab   No. spaces
column#   position   required
   1          9         8
   4          9         5
   8          9         1
  18         25         7

You should be able to see a pattern here. If we are already at a tab position (1, 9, 17, etc.), we need to add 8 spaces. And in general, if we are n characters past a tab position (where 0 <= n <= 7), then we need to add 8-n spaces.

We can calculate n easily:

n = (pos - 1) % 8

so nb (the number of spaces we need to add) can simply be calculated as follows:

nb = 8 - n

Hence

nb = 8 - ((pos - 1) % 8)

Or, more generally,

nb = TABINC - ((pos - 1) % TABINC)