Understanding the word count program from K&R book

I have started learning C programming using the classic C book by Ritchie & couple of other books as a reference. While going through solved examples I came across this program which counts new lines, characters & words. I understood new line and character part without any issues (obviously). However the logic to count words is confusing me. Can any one please help me in deciphering the code (in blue)?

Code:

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#include <stdio.h>
#include <stdlib.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* count lines, words, and characters in input */
int main(void) {
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("line:%d word:%d Character:%d\n", nl, nw, nc);
return(0);
}

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It's keeping track of two states, in a word and out of a word. As soon as a white space character is found state is set to out.

If the character is not white space and state is out, then we are in a word, state is set to in, and the word count variable is incremented.

Quote:

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Originally Posted by Subsonics

If the character is not white space and state is out, then we are in a word, state is set to in, and the word count variable is incremented.

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Ummm...thanks I understood that if the stream of characters contains either ' ', \n or \t then obviously it cannot be word & state is set to 0 (out). How is it establishing that a stream of characters contains no whitespace? Can you please explain the highlighted text in greater detail?

What is the role of this piece of code?

Code:

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else if (state == OUT)

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I am sorry, I know you've tried to explain me but I still have some doubts lingering!

White space is the stuff between words, and yes there are more white space characters than tab, space and new line but that is the basic idea.

The whole point of using this approach is to count only once per word. You asked about the check:

Code:

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else if (state == OUT)

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Well, that makes sure you can not take that code path more than once until a new white space is encountered. This, since state is set to IN, inside that block.

Quote:

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Originally Posted by alter.ego

Code:

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#include <stdio.h>
#include <stdlib.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* count lines, words, and characters in input */
int main(void) {
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("line:%d word:%d Character:%d\n", nl, nw, nc);
return(0);
}

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This is one case where text formatting helps a lot... if you indent and comment your code properly you can almost see what it's doing...

Code:

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#include <stdio.h>
#include <stdlib.h>

#define IN 1      // parsing a word
#define OUT 0    // not parsing a word

int main(void)
  {  int c = 0;      // character from keyboard
    int nl = 0;      // new line counter
    int nw = 0;      // number of words
    int nc = 0;      // number of characters
    int state = OUT; // initially not inside a word

    while ((c = getchar()) != EOF)
      { 
          ++nc;        // count characters   

          if (c == '\n')  // count lines
            ++nl;

          if (c == ' ' || c == '\n' || c == '\t')
            state = OUT;  // whitespace found
          else if (state == OUT)
            {
              state = IN;  // enter a word
              ++nw;        // count words
            }
      }

  // display summary after CTRL-Z
  printf("line:%d word:%d Character:%d\n", nl, nw, nc);
  return(0);
}

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Thanks for taking out time & helping me out guys! Those last few lines in the code are still bothering me. In the meanwhile, I used a twisted logic (no. of words = blank spaces + 1) to create a much simpler program.

Code:

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#include <stdio.h>
#include <stdlib.h>
/* #define IN 1 inside a word */
/* #define OUT 0 outside a word */
/* count lines, words, and characters in input */
int main(void) {
int c, nl, nw, nc, nb, state;
/*state = OUT;*/
nl = nw = nc = nb = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ')
++nb;
/*if (c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if (state == OUT)
{

        state = IN;
        ++nw;

}*/
}
printf("line:%d word:%d Character:%d\n", nl, nb+1, nc);
return(0);
}

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Its know it is silly and will fail if someone presses space-bar twice (among other scenarios) but at least it works in 'normal' circumstances.

Good stuff... now all you've got to learn is how to indent your code properly and toss in a few comments...

Think of it this way... You're working on a relatively complex project... you write it like you did above... if you come back to it in 5 years are you still going to be able to follow it?

You should cultivate Good habits early... rather than fixing things up after the bad ones, later.

Will def. cultivate this habit. Clumsiness of real life will not spill here :). Thanks!

Code:

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else if (state == OUT) {
state = IN;
++nw;

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Apologies to dig this old thread but another challenge brought me back. How does state = IN; signifies that now the control is in the middle of a word and that it should count it?

I mean when variable c encounters any wildchars is easier to determine that its not a word (its outside the word) but how does vice versa happens? I want to understand this aspect of programming wherein variables are used to keep track of states within the program. Any help will be greatly appreciated.

Ok, here's a dead simple example that does essentially the same thing as the IN and OUT in your exercise... but with a different purpose.

Code:

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#define On 1
#define Off 0

int IsOn = Off;

// bunches of code...

if (IsOn)  // test if the bulb is on
  { IsOn = Off;                // flag it as off
    TurnOffLamp(); }      // turn it off
else
  { IsOn = On;                // flag it as on
    TurnOnLamp(); }      // turn it on

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You will find flag variables of this kind are used quite often. The flag merely directs program flow... in this case to toggle a light bulb on and off ...

> while ((c = getchar()) != EOF)
Well ordinarily, state should be OUT when you exit this loop.
If the state it still IN, then perhaps you need to adjust the count in some way?

Quote:

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Originally Posted by CommonTater

Code:

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          else if (state == OUT)
            {
              state = IN;  // enter a word
              ++nw;        // count words
            }

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It says that if none of the cases above are true, then state is actually 'IN' a word.