is this simple program correct? K&R 1-12

This is how I solved the exercise 1-12 from K&R. I have found another solution online, but mine seems a lot easier than the other, even if sometimes it prints too many newlines.
Is this solution acceptable? If not, why? Thank you in advance.

FSX

Code:

---

#include <stdio.h>

int main(void)
{
        int c;

        while((c = getchar()) != EOF)
        {
                if((c == ' ') || (c == '\t'))
                        putchar('\n');
                else
                        putchar(c);
        }
        return 0;
}

---

Quote:

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Originally Posted by fsx

Is this solution acceptable? If not, why?

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Since your program is not a correct solution, it is especially not acceptable. Re-read the exercise: "Write a program that prints its input one word per line." Your program occasionally prints empty lines, i.e. lines that don't contain a word (and certainly K&R weren't thinking about the empty word here).

Try to get rid of these empty lines, e.g. by introducing two states "last character was a whitespace" and "last character wasn't". Also have a look at the isspace() function/macro.

Quote:

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mine seems a lot easier than the other, even if sometimes it prints too many newlines.

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If you allow your program to occasionally produce incorrect results, here's an even simpler version:

Code:

---

int main() {}

---

;-),
Philip

Quote:

---

Originally Posted by fsx

This is how I solved the exercise 1-12 from K&R. I have found another solution online, but mine seems a lot easier than the other, even if sometimes it prints too many newlines.
Is this solution acceptable? If not, why? Thank you in advance.

FSX

Code:

---

#include <stdio.h>

int main(void)
{
        int c;

        while((c = getchar()) != EOF)
        {
                if((c == ' ') || (c == '\t'))
                        putchar('\n');
                else
                        putchar(c);
        }
        return 0;
}

---

---

i executed ur code and it seems to me its working properly.i simply inputted a line and it gave me each word of the line in a new line.

Try inputting two or more lines, Ben. You'll see the code prints additional blank lines.

In any event, fsx, Snafuist is right - printing additional unwanted output hardly qualifies as a correct solution. The objective of writing code is to deliver the required functionality: no more, no less. Having code that is "easier" is not an excuse for delivering slightly different functionality. The programmer is expected to meet requirements, not force the user to accept compromised requirements.

Quote:

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Originally Posted by fsx

Code:

---

        while((c = getchar()) != EOF)
        {
                if((c == ' ') || (c == '\t'))
                        putchar('\n');

---

---

What you really want is a "nop" (no-op, no operation) and to continue --

Code:

---

                if((c == ' ') || (c == '\t'))
                        continue;

---

Except you still need a line break! So maybe

Code:

---

#include <stdio.h>

int main(void)
{
        int c;
        char flag=0;

        while((c = getchar()) != EOF)
        {
                if((c == ' ') || (c == '\t'))
                        if (flag) continue;
                        else { putchar('\n');
                                flag=1; }
                else {        putchar(c);
                        flag=0; }
        }
        return 0;
}

---

If you are just on chapter 1 you may not know about "continue" (it just starts the loop again without doing anything else).

The use of a flag is a very handy, basic, chapter 1 kind of thing, so if you don't understand what that's about, try to, or ask and I'll explain.

Basically now you could type:

Code:

---

this      is    a  sentance  with  extra      spaces
Output:
this
is
a
sentence
with
extra
spaces


---

That was the idea that Snafuist introduced at the beginning. I think you could come up with a better name than "flag" for that variable, though. Maybe something like "saw_whitespace". There's also a logical bug in the program (which was present in my first version too): since the flag is initially zero, there will be a blank line at the beginning of the output if the first character is whitespace. Also, continue isn't really necessary.

Anyway, here's my shot at it.

Code:

---

#include <stdio.h>
#include <ctype.h>

int main() {
    int c, between_words = 1;

    while((c = getchar()) != EOF) {
        if(isspace(c)) {
            if(!between_words) {
                putchar('\n');
                between_words = 1;
            }
        }
        else {
            putchar(c);
            between_words = 0;
        }
    }

    return 0;
}

---

Note the use of isspace() from ctype.h. isspace [C++ Reference]

I am still stumbling...

Code:

---

#include <stdio.h>

int main(void)
{
        int c, pc;

        while((c = getchar()) != EOF)
        {
                if((c == ' ') || (c == '\t') && (pc != c))
                        putchar('\n');
                else if((c != ' ') && (c != '\t'))
                        putchar(c);
                else
                        putchar(c);
                pc = c;
        }
        return 0;
}

---

What is not working?

I would expect that you will need a nested if-statement:

Code:

---

  if (c == ' ' || c == '\t')
  {
    if (pc != c)
      putchar('\n');
  }
  else
      putchar(c);

---

This will prevent multiple spaces from giving multiple newlines.

--
Mats

Thanks a lot! What I would like to understand is why if I use the && it doesn't works... shouldn't it be similar to another IF statement?

Quote:

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Originally Posted by matsp

What is not working?

I would expect that you will need a nested if-statement:

Code:

---

  if (c == ' ' || c == '\t')
  {
    if (pc != c)
      putchar('\n');
  }
  else
      putchar(c);

---

This will prevent multiple spaces from giving multiple newlines.

--
Mats

---

Because you do not want the ELSE part to be executed if the second if is false - only if the first if is false.

You could of course add a "if (pc != c)" inside the else-branch instead, but that's less logical.

--
Mats

Quote:

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Originally Posted by dwks

That was the idea that Snafuist introduced at the beginning. I think you could come up with a better name than "flag" for that variable, though.

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I called it "flag" because it is a (boolean) flag and I wanted the OP (fsx) to pick up on that (Snafuist refers to this as "two states", which is not quite as specific -- I guess you could see a boolean flag as an implementation of a two-state concept).

You're right about the continue being unnecessary and the initial whitespace issue.

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

Quote:

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Originally Posted by MK27

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

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yes it could. you need profiling to prove it or opposite (when the cache size is critical issue)

Quote:

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Originally Posted by matsp

Because you do not want the ELSE part to be executed if the second if is false - only if the first if is false.

You could of course add a "if (pc != c)" inside the else-branch instead, but that's less logical.

--
Mats

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Mats got exactly what I was looking for, a simple explanation. I'm new to C and all those pointers, includes and so on are totally alien to my current knowledge. But thanks also for the other answerers.

Quote:

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Originally Posted by MK27

Here's a related question: I usually use a char for boolean values cause it's smaller, but would it be true that an int is actually faster for the processor because it has a four byte boundary? Sort of nit-picking, but this could have a lot of significance in the right place...

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Can't bit fields be used for this (making the int size smaller) ? Or is the struct making it larger again?

Code:

---

#include<stdio.h>

struct foo
{
        int flag:2;
} bar;

int main(void)
{
        bar.flag = 1;
        printf("%d", bar.flag);
       
        return 0;
}

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Bitfields won't actually shrink the structure smaller than the actual type. That is to say, even if you set up a bit field of :2, it's still going to be at least an int in size.

Quzah.